{"id":410,"date":"2021-02-04T02:02:54","date_gmt":"2021-02-04T02:02:54","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=410"},"modified":"2022-03-16T05:49:12","modified_gmt":"2022-03-16T05:49:12","slug":"solving-optimization-problems","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/solving-optimization-problems\/","title":{"raw":"Solving Optimization Problems","rendered":"Solving Optimization Problems"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Set up and solve optimization problems in several applied fields<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1165042514973\" class=\"bc-section section\">\r\n<h2>Solving Optimization Problems over a Closed, Bounded Interval<\/h2>\r\n<p id=\"fs-id1165043329454\">The basic idea of the<strong> optimization problems<\/strong> that follow is the same. We have a particular quantity that we are interested in maximizing or minimizing. However, we also have some auxiliary condition that needs to be satisfied. For instance, in the example below, we are interested in maximizing the area of a rectangular garden. Certainly, if we keep making the side lengths of the garden larger, the area will continue to become larger. However, what if we have some restriction on how much fencing we can use for the perimeter? In this case, we cannot make the garden as large as we like. Let\u2019s look at how we can maximize the area of a rectangle subject to some constraint on the perimeter.<\/p>\r\n\r\n<div class=\"textbook exercises\">\r\n<h3>Example: Maximizing the Area of a Garden<\/h3>\r\n<p id=\"fs-id1165042623548\">A rectangular garden is to be constructed using a rock wall as one side of the garden and wire fencing for the other three sides (Figure 1). Given 100 ft of wire fencing, determine the dimensions that would create a garden of maximum area. What is the maximum area?<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"462\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211219\/CNX_Calc_Figure_04_07_001.jpg\" alt=\"A drawing of a garden has x and y written on the vertical and horizontal sides, respectively. There is a rock wall running along the entire bottom horizontal length of the drawing.\" width=\"462\" height=\"378\" \/> Figure 1. We want to determine the measurements [latex]x[\/latex] and [latex]y[\/latex] that will create a garden with a maximum area using 100 ft of fencing.[\/caption][reveal-answer q=\"fs-id1165043262577\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043262577\"]\r\n<p id=\"fs-id1165043262577\">Let [latex]x[\/latex] denote the length of the side of the garden perpendicular to the rock wall and [latex]y[\/latex] denote the length of the side parallel to the rock wall. Then the area of the garden is<\/p>\r\n\r\n<div id=\"fs-id1165042515463\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A=x\u00b7y[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043222797\">We want to find the maximum possible area subject to the constraint that the total fencing is [latex]100[\/latex] ft. From Figure 1, the total amount of fencing used will be [latex]2x+y[\/latex]. Therefore, the constraint equation is<\/p>\r\n\r\n<div id=\"fs-id1165043249986\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]2x+y=100[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043272012\">Solving this equation for [latex]y[\/latex], we have [latex]y=100-2x[\/latex]. Thus, we can write the area as<\/p>\r\n\r\n<div id=\"fs-id1165043281405\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A(x)=x \\cdot (100-2x)=100x-2x^2[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043195368\">Before trying to maximize the area function [latex]A(x)=100x-2x^2[\/latex], we need to determine the domain under consideration. To construct a rectangular garden, we certainly need the lengths of both sides to be positive. Therefore, we need [latex]x&gt;0[\/latex] and [latex]y&gt;0[\/latex]. Since [latex]y=100-2x[\/latex], if [latex]y&gt;0[\/latex], then [latex]x&lt;50[\/latex]. Therefore, we are trying to determine the maximum value of [latex]A(x)[\/latex] for [latex]x[\/latex] over the open interval [latex](0,50)[\/latex]. We do not know that a function necessarily has a maximum value over an open interval. However, we do know that a continuous function has an absolute maximum (and absolute minimum) over a closed interval. Therefore, let\u2019s consider the function [latex]A(x)=100x-2x^2[\/latex] over the closed interval [latex][0,50][\/latex]. If the maximum value occurs at an interior point, then we have found the value [latex]x[\/latex] in the open interval [latex](0,50)[\/latex] that maximizes the area of the garden. Therefore, we consider the following problem:<\/p>\r\n<p id=\"fs-id1165043094646\" style=\"text-align: center;\">Maximize [latex]A(x)=100x-2x^2[\/latex] over the interval [latex][0,50][\/latex]<\/p>\r\n<p id=\"fs-id1165042989293\">As mentioned earlier, since [latex]A[\/latex] is a continuous function on a closed, bounded interval, by the extreme value theorem, it has a maximum and a minimum. These extreme values occur either at endpoints or critical points. At the endpoints, [latex]A(x)=0[\/latex]. Since the area is positive for all [latex]x[\/latex] in the open interval [latex](0,50)[\/latex], the maximum must occur at a critical point. Differentiating the function [latex]A(x)[\/latex], we obtain<\/p>\r\n\r\n<div id=\"fs-id1165043380052\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A^{\\prime}(x)=100-4x[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043178466\">Therefore, the only critical point is [latex]x=25[\/latex] (Figure 2). We conclude that the maximum area must occur when [latex]x=25[\/latex]. Then we have [latex]y=100-2x=100-2(25)=50[\/latex]. To maximize the area of the garden, let [latex]x=25[\/latex] ft and [latex]y=50[\/latex] ft. The area of this garden is [latex]1250 \\, \\text{ft}^2[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"534\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211223\/CNX_Calc_Figure_04_07_002.jpg\" alt=\"The function A(x) = 100x \u2013 2x is graphed. At its maximum there is an intersection of two dashed lines and text that reads \u201cMaximum area is 1250 square feet when x = 25 feet.\u201d\" width=\"534\" height=\"347\" \/> Figure 2. To maximize the area of the garden, we need to find the maximum value of the function [latex]A(x)=100x-2x^2[\/latex].[\/caption][\/hidden-answer]<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Maximizing the Area of a Garden.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/_lNDHOV2FlQ?controls=0&amp;start=15&amp;end=180&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.7AppliedOptimizationProblems15to180_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.7 Applied Optimization Problems\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1165042474337\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165043035196\">Determine the maximum area if we want to make the same rectangular garden as in Figure 2, but we have 200 ft of fencing.<\/p>\r\n[reveal-answer q=\"355002\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"355002\"]\r\n\r\nWe need to maximize the function [latex]A(x)=200x-2x^2[\/latex] over the interval [latex][0,100][\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165043102115\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043102115\"]\r\n<p id=\"fs-id1165043102115\">The maximum area is [latex]5000 \\, \\text{ft}^2[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165042376423\">Now let\u2019s look at a general strategy for solving optimization problems similar to these above.<\/p>\r\n\r\n<div id=\"fs-id1165042916977\" class=\"textbox examples\">\r\n<h3>Problem-Solving Strategy: Solving Optimization Problems<\/h3>\r\n<ol id=\"fs-id1165043013367\">\r\n \t<li>Introduce all variables. If applicable, draw a figure and label all variables.<\/li>\r\n \t<li>Determine which quantity is to be maximized or minimized, and for what range of values of the other variables (if this can be determined at this time).<\/li>\r\n \t<li>Write a formula for the quantity to be maximized or minimized in terms of the variables. This formula may involve more than one variable.<\/li>\r\n \t<li>Write any equations relating the independent variables in the formula from step 3. Use these equations to write the quantity to be maximized or minimized as a function of one variable.<\/li>\r\n \t<li>Identify the domain of consideration for the function in step 4 based on the physical problem to be solved.<\/li>\r\n \t<li>Locate the maximum or minimum value of the function from step 4. This step typically involves looking for critical points and evaluating a function at endpoints.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<p id=\"fs-id1165043029463\">Now let\u2019s apply this strategy to maximize the volume of an open-top box given a constraint on the amount of material to be used.<\/p>\r\n\r\n<div id=\"fs-id1165043024669\" class=\"textbook exercises\">\r\n<h3>example: Maximizing the Volume of a Box<\/h3>\r\n<p id=\"fs-id1165042638735\">An open-top box is to be made from a 24 in. by 36 in. piece of cardboard by removing a square from each corner of the box and folding up the flaps on each side. What size square should be cut out of each corner to get a box with the maximum volume?<\/p>\r\n[reveal-answer q=\"fs-id1165043092635\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043092635\"]\r\n<p id=\"fs-id1165043092635\">Step 1: Let [latex]x[\/latex] be the side length of the square to be removed from each corner (Figure 3). Then, the remaining four flaps can be folded up to form an open-top box. Let [latex]V[\/latex] be the volume of the resulting box.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"599\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211226\/CNX_Calc_Figure_04_07_003.jpg\" alt=\"There are two figures for this figure. The first one is a rectangle with sides 24 in and 36 in, with each corner having a square of side length x taken out of it. In the second picture, there is a box with side lengths x in, 24 \u2013 2x in, and 36 \u2013 2x in.\" width=\"599\" height=\"333\" \/> Figure 3. A square with side length [latex]x[\/latex] inches is removed from each corner of the piece of cardboard. The remaining flaps are folded to form an open-top box.[\/caption]\r\n<p id=\"fs-id1165042330556\">Step 2: We are trying to maximize the volume of a box. Therefore, the problem is to maximize [latex]V[\/latex].<\/p>\r\n<p id=\"fs-id1165043251052\">Step 3: As mentioned in step 2, are trying to maximize the volume of a box. The volume of a box is [latex]V=L \\cdot W \\cdot H[\/latex], where [latex]L, \\, W[\/latex], and [latex]H[\/latex] are the length, width, and height, respectively.<\/p>\r\nStep 4: From Figure 3, we see that the height of the box is [latex]x[\/latex] inches, the length is [latex]36-2x[\/latex] inches, and the width is [latex]24-2x[\/latex] inches. Therefore, the volume of the box is\r\n<div id=\"fs-id1165042645725\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V(x)=(36-2x)(24-2x)x=4x^3-120x^2+864x[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043115396\">Step 5: To determine the domain of consideration, let\u2019s examine Figure 3. Certainly, we need [latex]x&gt;0[\/latex]. Furthermore, the side length of the square cannot be greater than or equal to half the length of the shorter side, 24 in.; otherwise, one of the flaps would be completely cut off. Therefore, we are trying to determine whether there is a maximum volume of the box for [latex]x[\/latex] over the open interval [latex](0,12)[\/latex]. Since [latex]V[\/latex] is a continuous function over the closed interval [latex][0,12][\/latex], we know [latex]V[\/latex] will have an absolute maximum over the closed interval. Therefore, we consider [latex]V[\/latex] over the closed interval [latex][0,12][\/latex] and check whether the absolute maximum occurs at an interior point.<\/p>\r\n<p id=\"fs-id1165043017956\">Step 6: Since [latex]V(x)[\/latex] is a continuous function over the closed, bounded interval [latex][0,12][\/latex], [latex]V[\/latex] must have an absolute maximum (and an absolute minimum). Since [latex]V(x)=0[\/latex] at the endpoints and [latex]V(x)&gt;0[\/latex] for [latex]0&lt;x&lt;12[\/latex], the maximum must occur at a critical point. The derivative is<\/p>\r\n\r\n<div id=\"fs-id1165043333862\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V^{\\prime}(x)=12x^2-240x+864[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042608183\">To find the critical points, we need to solve the equation<\/p>\r\n\r\n<div id=\"fs-id1165042946835\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]12x^2-240x+864=0[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043209010\">Dividing both sides of this equation by 12, the problem simplifies to solving the equation<\/p>\r\n\r\n<div id=\"fs-id1165042982075\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x^2-20x+72=0[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043060298\">Using the quadratic formula, we find that the critical points are<\/p>\r\n\r\n<div id=\"fs-id1165042367652\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x=\\frac{20 \\pm \\sqrt{(-20)^2-4(1)(72)}}{2}=\\frac{20 \\pm \\sqrt{112}}{2}=\\frac{20 \\pm 4\\sqrt{7}}{2}=10 \\pm 2\\sqrt{7}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043263257\">Since [latex]10+2\\sqrt{7}[\/latex] is not in the domain of consideration, the only critical point we need to consider is [latex]10-2\\sqrt{7}[\/latex]. Therefore, the volume is maximized if we let [latex]x=10-2\\sqrt{7}[\/latex] in. The maximum volume is [latex]V(10-2\\sqrt{7})=640+448\\sqrt{7}\\approx 1825 \\, \\text{in}^3[\/latex] as shown in the following graph.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"719\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211230\/CNX_Calc_Figure_04_07_004.jpg\" alt=\"The function V(x) = 4x3 \u2013 120x2 + 864x is graphed. At its maximum there is an intersection of two dashed lines and text that reads \u201cMaximum volume is approximately 1825 cubic inches when x \u2248 4.7 inches.\u201d\" width=\"719\" height=\"388\" \/> Figure 4. Maximizing the volume of the box leads to finding the maximum value of a cubic polynomial.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165043019959\" class=\"textbox tryit\">\r\n<h3>Interactive<\/h3>\r\n<a href=\"https:\/\/www.khanacademy.org\/math\/old-differential-calculus\/derivative-applications-dc\/optimization-dc\/v\/optimizing-box-volume-graphically\" target=\"_blank\" rel=\"noopener\">Watch a video about optimizing the volume of a box.<\/a>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165042992008\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165043108738\">Suppose the dimensions of the cardboard in the previous example are 20 in. by 30 in. Let [latex]x[\/latex] be the side length of each square and write the volume of the open-top box as a function of [latex]x[\/latex]. Determine the domain of consideration for [latex]x[\/latex].<\/p>\r\n[reveal-answer q=\"8044661\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"8044661\"]\r\n<p id=\"fs-id1165043080382\">The volume of the box is [latex]L \\cdot W \\cdot H[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165043177823\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043177823\"]\r\n<p id=\"fs-id1165043177823\">[latex]V(x)=x(20-2x)(30-2x)[\/latex]. The domain is [latex][0,10][\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165042350080\" class=\"textbook exercises\">\r\n<h3>Example: Minimizing Travel Time<\/h3>\r\n<p id=\"fs-id1165043329514\">An island is [latex]2[\/latex] mi due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that is [latex]6[\/latex] mi west of that point. The visitor is planning to go from the cabin to the island. Suppose the visitor runs at a rate of [latex]8[\/latex] mph and swims at a rate of [latex]3[\/latex] mph. How far should the visitor run before swimming to minimize the time it takes to reach the island?<\/p>\r\n[reveal-answer q=\"fs-id1165043276181\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043276181\"]\r\n<p id=\"fs-id1165043276181\">Step 1: Let [latex]x[\/latex] be the distance running and let [latex]y[\/latex] be the distance swimming (Figure 5). Let [latex]T[\/latex] be the time it takes to get from the cabin to the island.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"594\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211233\/CNX_Calc_Figure_04_07_011.jpg\" alt=\"The cabin is x miles from the shore. From that point on the shore, the island is y miles away. If you were to continue the line from the cabin to the shore (the x miles one) and if you were to draw a line from the island parallel to the shore, then the lines would extend 2 miles from the island and 6 miles from the cabin before intersecting.\" width=\"594\" height=\"352\" \/> Figure 5. How can we choose [latex]x[\/latex] and [latex]y[\/latex] to minimize the travel time from the cabin to the island?[\/caption]\r\n<p id=\"fs-id1165042330905\">Step 2: The problem is to minimize [latex]T[\/latex].<\/p>\r\n<p id=\"fs-id1165043346985\">Step 3: To find the time spent traveling from the cabin to the island, add the time spent running and the time spent swimming. Since Distance [latex]=[\/latex] Rate [latex]\\times[\/latex] Time [latex](D=R \\times T)[\/latex], the time spent running is<\/p>\r\n\r\n<div id=\"fs-id1165043431452\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T_{\\text{running}}=\\frac{D_{\\text{running}}}{R_{\\text{running}}}=\\frac{x}{8}[\/latex],<\/div>\r\nand the time spent swimming is\r\n<div id=\"fs-id1165042369650\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T_{\\text{swimming}}=\\frac{D_{\\text{swimming}}}{R_{\\text{swimming}}}=\\frac{y}{3}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043262597\">Therefore, the total time spent traveling is<\/p>\r\n\r\n<div id=\"fs-id1165043433402\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T=\\frac{x}{8}+\\frac{y}{3}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042517820\">Step 4: From Figure 5, the line segment of [latex]y[\/latex] miles forms the hypotenuse of a right triangle with legs of length [latex]2[\/latex] mi and [latex]6-x[\/latex] mi. Therefore, by the Pythagorean theorem, [latex]2^2+(6-x)^2=y^2[\/latex], and we obtain [latex]y=\\sqrt{(6-x)^2+4}[\/latex]. Thus, the total time spent traveling is given by the function<\/p>\r\n\r\n<div id=\"fs-id1165043430018\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T(x)=\\frac{x}{8}+\\frac{\\sqrt{(6-x)^2+4}}{3}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042328992\">Step 5: From Figure 5, we see that [latex]0\\le x\\le 6[\/latex]. Therefore, [latex][0,6][\/latex] is the domain of consideration.<\/p>\r\n<p id=\"fs-id1165043066617\">Step 6: Since [latex]T(x)[\/latex] is a continuous function over a closed, bounded interval, it has a maximum and a minimum. Let\u2019s begin by looking for any critical points of [latex]T[\/latex] over the interval [latex][0,6][\/latex]. The derivative is<\/p>\r\n\r\n<div id=\"fs-id1165043353795\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T^{\\prime}(x)=\\frac{1}{8}-\\frac{1}{2}\\frac{[(6-x)^2+4]^{-1\/2}}{3} \\cdot 2(6-x)=\\frac{1}{8}-\\frac{(6-x)}{3\\sqrt{(6-x)^2+4}}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042516379\">If [latex]T^{\\prime}(x)=0[\/latex], then<\/p>\r\n\r\n<div id=\"fs-id1165042705986\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{8}=\\frac{6-x}{3\\sqrt{(6-x)^2+4}}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043107403\">Therefore,<\/p>\r\n\r\n<div id=\"fs-id1165042343282\" class=\"equation\" style=\"text-align: center;\">[latex]3\\sqrt{(6-x)^2+4}=8(6-x)[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043284394\">Squaring both sides of this equation, we see that if [latex]x[\/latex] satisfies this equation, then [latex]x[\/latex] must satisfy<\/p>\r\n\r\n<div id=\"fs-id1165043103774\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]9[(6-x)^2+4]=64(6-x)^2[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042923339\">which implies<\/p>\r\n\r\n<div id=\"fs-id1165043285288\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]55(6-x)^2=36[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042711030\">We conclude that if [latex]x[\/latex] is a critical point, then [latex]x[\/latex] satisfies<\/p>\r\n\r\n<div id=\"fs-id1165042923510\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](x-6)^2=\\frac{36}{55}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043396085\">Therefore, the possibilities for critical points are<\/p>\r\n\r\n<div id=\"fs-id1165042350053\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x=6 \\pm \\frac{6}{\\sqrt{55}}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042616899\">Since [latex]x=6+\\frac{6}{\\sqrt{55}}[\/latex] is not in the domain, it is not a possibility for a critical point. On the other hand, [latex]x=6-\\frac{6}{\\sqrt{55}}[\/latex] is in the domain. Since we squared both sides of the equation to arrive at the possible critical points, it remains to verify that it is satisfied by [latex]x=6-\\frac{6}{\\sqrt{55}}[\/latex]. Since [latex]x=6-\\frac{6}{\\sqrt{55}}[\/latex] does satisfy that equation, we conclude that [latex]x=6-\\frac{6}{\\sqrt{55}}[\/latex] is a critical point, and it is the only one. To justify that the time is minimized for this value of [latex]x[\/latex], we just need to check the values of [latex]T(x)[\/latex] at the endpoints [latex]x=0[\/latex] and [latex]x=6[\/latex], and compare them with the value of [latex]T(x)[\/latex] at the critical point [latex]x=6-\\frac{6}{\\sqrt{55}}[\/latex]. We find that [latex]T(0)\\approx 2.108[\/latex] h and [latex]T(6)\\approx 1.417[\/latex] h, whereas [latex]T(6-\\frac{6}{\\sqrt{55}})\\approx 1.368[\/latex] h. Therefore, we conclude that [latex]T[\/latex] has a local minimum at [latex]x\\approx 5.19[\/latex] mi.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165043107415\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165043348922\">Suppose the island is 1 mi from shore, and the distance from the cabin to the point on the shore closest to the island is [latex]15[\/latex] mi. Suppose a visitor swims at the rate of [latex]2.5[\/latex] mph and runs at a rate of [latex]6[\/latex] mph. Let [latex]x[\/latex] denote the distance the visitor will run before swimming, and find a function for the time it takes the visitor to get from the cabin to the island.<\/p>\r\n[reveal-answer q=\"212787\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"212787\"]\r\n\r\nThe time [latex]T=T_{\\text{running}}+T_{\\text{swimming}}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165042369705\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042369705\"]\r\n<p id=\"fs-id1165042369705\">[latex]T(x)=\\frac{x}{6}+\\frac{\\sqrt{(15-x)^2+1}}{2.5}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165042556259\">In business, companies are interested in <span class=\"no-emphasis\">maximizing revenue<\/span>. In the following example, we consider a scenario in which a company has collected data on how many cars it is able to lease, depending on the price it charges its customers to rent a car. Let\u2019s use these data to determine the price the company should charge to maximize the amount of money it brings in.<\/p>\r\n\r\n<div id=\"fs-id1165043353574\" class=\"textbook exercises\">\r\n<h3>Example: Maximizing Revenue<\/h3>\r\n<p id=\"fs-id1165042515832\">Owners of a car rental company have determined that if they charge customers [latex]p[\/latex] dollars per day to rent a car, where [latex]50\\le p\\le 200[\/latex], the number of cars [latex]n[\/latex] they rent per day can be modeled by the linear function [latex]n(p)=1000-5p[\/latex]. If they charge [latex]$50[\/latex] per day or less, they will rent all their cars. If they charge [latex]$200[\/latex] per day or more, they will not rent any cars. Assuming the owners plan to charge customers between [latex]$50[\/latex] per day and [latex]$200[\/latex] per day to rent a car, how much should they charge to maximize their revenue?<\/p>\r\n[reveal-answer q=\"fs-id1165043423441\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043423441\"]\r\n<p id=\"fs-id1165043423441\">Step 1: Let [latex]p[\/latex] be the price charged per car per day and let [latex]n[\/latex] be the number of cars rented per day. Let [latex]R[\/latex] be the revenue per day.<\/p>\r\n<p id=\"fs-id1165043431290\">Step 2: The problem is to maximize [latex]R[\/latex].<\/p>\r\n<p id=\"fs-id1165043087644\">Step 3: The revenue (per day) is equal to the number of cars rented per day times the price charged per car per day\u2014that is, [latex]R=n \\times p[\/latex].<\/p>\r\n<p id=\"fs-id1165042514435\">Step 4: Since the number of cars rented per day is modeled by the linear function [latex]n(p)=1000-5p[\/latex], the revenue [latex]R[\/latex] can be represented by the function<\/p>\r\n\r\n<div id=\"fs-id1165043256835\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]R(p)=n \\times p=(1000-5p)p=-5p^2+1000p[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042406610\">Step 5: Since the owners plan to charge between [latex]$50[\/latex] per car per day and [latex]$200[\/latex] per car per day, the problem is to find the maximum revenue [latex]R(p)[\/latex] for [latex]p[\/latex] in the closed interval [latex][50,200][\/latex].<\/p>\r\n<p id=\"fs-id1165043173962\">Step 6: Since [latex]R[\/latex] is a continuous function over the closed, bounded interval [latex][50,200][\/latex], it has an absolute maximum (and an absolute minimum) in that interval. To find the maximum value, look for critical points. The derivative is [latex]R^{\\prime}(p)=-10p+1000[\/latex]. Therefore, the critical point is [latex]p=100[\/latex] When [latex]p=100[\/latex], [latex]R(100)=$50,000[\/latex]. When [latex]p=50[\/latex], [latex]R(p)=$37,500[\/latex]. When [latex]p=200[\/latex], [latex]R(p)=$0[\/latex]. Therefore, the absolute maximum occurs at [latex]p=$100[\/latex]. The car rental company should charge [latex]$100[\/latex] per day per car to maximize revenue as shown in the following figure.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"495\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211236\/CNX_Calc_Figure_04_07_010.jpg\" alt=\"The function R(p) is graphed. At its maximum there is an intersection of two dashed lines and text that reads \u201cMaximum revenue is 50,000 per day when the price charged per car is 100 per day.\u201d\" width=\"495\" height=\"270\" \/> Figure 6. To maximize revenue, a car rental company has to balance the price of a rental against the number of cars people will rent at that price.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Maximizing Revenue.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/_lNDHOV2FlQ?controls=0&amp;start=913&amp;end=1024&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.7AppliedOptimizationProblems913to1024_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.7 Applied Optimization Problems\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1165043257501\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165042376530\">A car rental company charges its customers [latex]p[\/latex] dollars per day, where [latex]60\\le p\\le 150[\/latex]. It has found that the number of cars rented per day can be modeled by the linear function [latex]n(p)=750-5p[\/latex]. How much should the company charge each customer to maximize revenue?<\/p>\r\n[reveal-answer q=\"13435467\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"13435467\"]\r\n<p id=\"fs-id1165042321515\">[latex]R(p)=n \\times p[\/latex], where [latex]n[\/latex] is the number of cars rented and [latex]p[\/latex] is the price charged per car.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165042960136\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042960136\"]\r\n<p id=\"fs-id1165042960136\">The company should charge [latex]$75[\/latex] per car per day.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165043350161\" class=\"textbook exercises\">\r\n<h3>Example: Maximizing the Area of an Inscribed Rectangle<\/h3>\r\n<p id=\"fs-id1165042456921\">A rectangle is to be inscribed in the ellipse<\/p>\r\n\r\n<div id=\"fs-id1165042456924\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{x^2}{4}+y^2=1[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042645676\">What should the dimensions of the rectangle be to maximize its area? What is the maximum area?<\/p>\r\n[reveal-answer q=\"fs-id1165042519826\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042519826\"]\r\n<p id=\"fs-id1165042519826\">Step 1: For a rectangle to be inscribed in the ellipse, the sides of the rectangle must be parallel to the axes. Let [latex]L[\/latex] be the length of the rectangle and [latex]W[\/latex] be its width. Let [latex]A[\/latex] be the area of the rectangle.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"575\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211239\/CNX_Calc_Figure_04_07_007.jpg\" alt=\"The ellipse x2\/4 + y2 = 1 is drawn with its x intercepts being \u00b12 and its y intercepts being \u00b11. There is a rectangle inscribed in the ellipse with length L (in the x-direction) and width W.\" width=\"575\" height=\"350\" \/> Figure 7. We want to maximize the area of a rectangle inscribed in an ellipse.[\/caption]\r\n<p id=\"fs-id1165042371013\">Step 2: The problem is to maximize [latex]A[\/latex].<\/p>\r\n<p id=\"fs-id1165042377481\">Step 3: The area of the rectangle is [latex]A=L \\times W[\/latex].<\/p>\r\n<p id=\"fs-id1165043257695\">Step 4: Let [latex](x,y)[\/latex] be the corner of the rectangle that lies in the first quadrant, as shown in Figure 7. We can write length [latex]L=2x[\/latex] and width [latex]W=2y[\/latex]. Since [latex]\\frac{x^2}{4}+y^2=1[\/latex] and [latex]y&gt;0[\/latex], we have [latex]y=\\sqrt{1 - \\frac{x^2}{4}}[\/latex]. Therefore, the area is<\/p>\r\n\r\n<div id=\"fs-id1165042320390\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A=L \\times W=(2x)(2y)=4x\\sqrt{1 - \\frac{x^2}{4}}=2x\\sqrt{4-x^2}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043380076\">Step 5: From Figure 7, we see that to inscribe a rectangle in the ellipse, the [latex]x[\/latex]-coordinate of the corner in the first quadrant must satisfy [latex]0&lt;x&lt;2[\/latex]. Therefore, the problem reduces to looking for the maximum value of [latex]A(x)[\/latex] over the open interval [latex](0,2)[\/latex]. Since [latex]A(x)[\/latex] will have an absolute maximum (and absolute minimum) over the closed interval [latex][0,2][\/latex], we consider [latex]A(x)=2x\\sqrt{4-x^2}[\/latex] over the interval [latex][0,2][\/latex]. If the absolute maximum occurs at an interior point, then we have found an absolute maximum in the open interval.<\/p>\r\n<p id=\"fs-id1165042331466\">Step 6: As mentioned earlier, [latex]A(x)[\/latex] is a continuous function over the closed, bounded interval [latex][0,2][\/latex]. Therefore, it has an absolute maximum (and absolute minimum). At the endpoints [latex]x=0[\/latex] and [latex]x=2[\/latex], [latex]A(x)=0[\/latex]. For [latex]0&lt;x&lt;2[\/latex], [latex]A(x)&gt;0[\/latex]. Therefore, the maximum must occur at a critical point. Taking the derivative of [latex]A(x)[\/latex], we obtain<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} A^{\\prime}(x) &amp; =2\\sqrt{4-x^2}+2x \\cdot \\frac{1}{2\\sqrt{4-x^2}}(-2x) \\\\ &amp; =2\\sqrt{4-x^2}-\\frac{2x^2}{\\sqrt{4-x^2}} \\\\ &amp; =\\frac{8-4x^2}{\\sqrt{4-x^2}} \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043379942\">To find critical points, we need to find where [latex]A^{\\prime}(x)=0[\/latex]. We can see that if [latex]x[\/latex] is a solution of<\/p>\r\n\r\n<div id=\"fs-id1165043354675\" class=\"equation\" style=\"text-align: center;\">[latex]\\frac{8-4x^2}{\\sqrt{4-x^2}}=0[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043183716\">then [latex]x[\/latex] must satisfy<\/p>\r\n\r\n<div id=\"fs-id1165043281800\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]8-4x^2=0[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042515040\">Therefore, [latex]x^2=2[\/latex]. Thus, [latex]x=\\pm \\sqrt{2}[\/latex] are the possible solutions of[latex]A^{\\prime}(x)[\/latex]. Since we are considering [latex]x[\/latex] over the interval [latex][0,2][\/latex], [latex]x=\\sqrt{2}[\/latex] is a possibility for a critical point, but [latex]x=\u2212\\sqrt{2}[\/latex] is not. Therefore, we check whether [latex]\\sqrt{2}[\/latex] is a solution of [latex]A^{\\prime}(x)[\/latex]. Since [latex]x=\\sqrt{2}[\/latex] is a solution of[latex]A^{\\prime}(x)[\/latex], we conclude that [latex]\\sqrt{2}[\/latex] is the only critical point of [latex]A(x)[\/latex] in the interval [latex][0,2][\/latex]. Therefore, [latex]A(x)[\/latex] must have an absolute maximum at the critical point [latex]x=\\sqrt{2}[\/latex]. To determine the dimensions of the rectangle, we need to find the length [latex]L[\/latex] and the width [latex]W[\/latex]. If [latex]x=\\sqrt{2}[\/latex] then<\/p>\r\n\r\n<div id=\"fs-id1165042708515\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\sqrt{1-\\frac{(\\sqrt{2})^2}{4}}=\\sqrt{1-\\frac{1}{2}}=\\frac{1}{\\sqrt{2}}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043376531\">Therefore, the dimensions of the rectangle are [latex]L=2x=2\\sqrt{2}[\/latex] and [latex]W=2y=\\frac{2}{\\sqrt{2}}=\\sqrt{2}[\/latex]. The area of this rectangle is [latex]A=L \\times W=(2\\sqrt{2})(\\sqrt{2})=4[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165043390986\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165043199268\">Modify the area function [latex]A[\/latex] if the rectangle is to be inscribed in the unit circle [latex]x^2+y^2=1[\/latex]. What is the domain of consideration?<\/p>\r\n[reveal-answer q=\"70845661\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"70845661\"]\r\n<p id=\"fs-id1165042318676\">If [latex](x,y)[\/latex] is the vertex of the square that lies in the first quadrant, then the area of the square is [latex]A=(2x)(2y)=4xy[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165043276112\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043276112\"]\r\n<p id=\"fs-id1165043276112\">[latex]A(x)=4x\\sqrt{1-x^2}[\/latex]. The domain of consideration is [latex][0,1][\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]54694[\/ohm_question]\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">Solving Optimization Problems when the Interval Is Not Closed or Is Unbounded<\/h2>\r\n<p id=\"fs-id1165042318512\">In the previous examples, we considered functions on closed, bounded domains. Consequently, by the extreme value theorem, we were guaranteed that the functions had absolute extrema. Let\u2019s now consider functions for which the domain is neither closed nor bounded.<\/p>\r\n<p id=\"fs-id1165043173977\">Many functions still have at least one absolute extrema, even if the domain is not closed or the domain is unbounded. For example, the function [latex]f(x)=x^2+4[\/latex] over [latex](\u2212\\infty ,\\infty )[\/latex] has an absolute minimum of 4 at [latex]x=0[\/latex]. Therefore, we can still consider functions over unbounded domains or open intervals and determine whether they have any absolute extrema. In the next example, we try to minimize a function over an unbounded domain. We will see that, although the domain of consideration is [latex](0,\\infty )[\/latex], the function has an absolute minimum.<\/p>\r\n<p id=\"fs-id1165042610536\">In the following example, we look at constructing a box of least surface area with a prescribed volume. It is not difficult to show that for a closed-top box, by symmetry, among all boxes with a specified volume, a cube will have the smallest surface area. Consequently, we consider the modified problem of determining which open-topped box with a specified volume has the smallest surface area.<\/p>\r\n\r\n<div id=\"fs-id1165042327673\" class=\"textbook exercises\">\r\n<h3>Example: Minimizing Surface Area<\/h3>\r\n<p id=\"fs-id1165043430067\">A rectangular box with a square base, an open top, and a volume of [latex]216 \\, \\text{in}^3[\/latex]\u00a0is to be constructed. What should the dimensions of the box be to minimize the surface area of the box? What is the minimum surface area?<\/p>\r\n[reveal-answer q=\"fs-id1165042373482\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042373482\"]\r\n<p id=\"fs-id1165042373482\">Step 1: Draw a rectangular box and introduce the variable [latex]x[\/latex] to represent the length of each side of the square base; let [latex]y[\/latex] represent the height of the box. Let [latex]S[\/latex] denote the surface area of the open-top box.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"281\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211241\/CNX_Calc_Figure_04_07_008.jpg\" alt=\"A box with square base is shown. The base has side length x, and the height is y.\" width=\"281\" height=\"264\" \/> Figure 8. We want to minimize the surface area of a square-based box with a given volume.[\/caption]\r\n<p id=\"fs-id1165042355751\">Step 2: We need to minimize the surface area. Therefore, we need to minimize [latex]S[\/latex].<\/p>\r\n<p id=\"fs-id1165043430591\">Step 3: Since the box has an open top, we need only determine the area of the four vertical sides and the base. The area of each of the four vertical sides is [latex]x \\cdot y[\/latex]. The area of the base is [latex]x^2[\/latex]. Therefore, the surface area of the box is<\/p>\r\n\r\n<div id=\"fs-id1165042327660\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]S=4xy+x^2[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042322398\">Step 4: Since the volume of this box is [latex]x^2 y[\/latex] and the volume is given as [latex]216 \\, \\text{in}^3[\/latex], the constraint equation is<\/p>\r\n\r\n<div id=\"fs-id1165042948321\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x^2 y=216[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042418854\">Solving the constraint equation for [latex]y[\/latex], we have [latex]y=\\frac{216}{x^2}[\/latex]. Therefore, we can write the surface area as a function of [latex]x[\/latex] only:<\/p>\r\n\r\n<div id=\"fs-id1165042375810\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]S(x)=4x(\\frac{216}{x^2})+x^2[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042318390\">Therefore, [latex]S(x)=\\frac{864}{x}+x^2[\/latex].<\/p>\r\n<p id=\"fs-id1165043393971\">Step 5: Since we are requiring that [latex]x^2 y=216[\/latex], we cannot have [latex]x=0[\/latex]. Therefore, we need [latex]x&gt;0[\/latex]. On the other hand, [latex]x[\/latex] is allowed to have any positive value. Note that as [latex]x[\/latex] becomes large, the height of the box [latex]y[\/latex] becomes correspondingly small so that [latex]x^2 y=216[\/latex]. Similarly, as [latex]x[\/latex] becomes small, the height of the box becomes correspondingly large. We conclude that the domain is the open, unbounded interval [latex](0,\\infty )[\/latex]. Note that, unlike the previous examples, we cannot reduce our problem to looking for an absolute maximum or absolute minimum over a closed, bounded interval. However, in the next step, we discover why this function must have an absolute minimum over the interval [latex](0,\\infty )[\/latex].<\/p>\r\n<p id=\"fs-id1165042638460\">Step 6: Note that as [latex]x\\to 0^+[\/latex], [latex]S(x)\\to \\infty[\/latex]. Also, as [latex]x\\to \\infty[\/latex], [latex]S(x)\\to \\infty[\/latex]. Since [latex]S[\/latex] is a continuous function that approaches infinity at the ends, it must have an absolute minimum at some [latex]x\\in (0,\\infty )[\/latex]. This minimum must occur at a critical point of [latex]S[\/latex]. The derivative is<\/p>\r\n\r\n<div id=\"fs-id1165043430791\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]S^{\\prime}(x)=-\\frac{864}{x^2}+2x[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042355317\">Therefore, [latex]S^{\\prime}(x)=0[\/latex] when [latex]2x=\\frac{864}{x^2}[\/latex]. Solving this equation for [latex]x[\/latex], we obtain [latex]x^3=432[\/latex], so [latex]x=\\sqrt[3]{432}=6\\sqrt[3]{2}[\/latex]. Since this is the only critical point of [latex]S[\/latex], the absolute minimum must occur at [latex]x=6\\sqrt[3]{2}[\/latex] (see Figure 9). When [latex]x=6\\sqrt[3]{2}[\/latex], [latex]y=\\frac{216}{(6\\sqrt[3]{2})^2}=3\\sqrt[3]{2}[\/latex] in. Therefore, the dimensions of the box should be [latex]x=6\\sqrt[3]{2}[\/latex] in and [latex]y=3\\sqrt[3]{2}[\/latex] in. With these dimensions, the surface area is<\/p>\r\n\r\n<div id=\"fs-id1165043379912\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]S(6\\sqrt[3]{2})=\\frac{864}{6\\sqrt[3]{2}}+(6\\sqrt[3]{2})^2=108\\sqrt[3]{4} \\, \\text{in}^2[\/latex]<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<div>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"477\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211244\/CNX_Calc_Figure_04_07_009.jpg\" alt=\"The function S(x) = 864\/x + x2 is graphed. At its minimum there is a dashed line and text that reads \u201cMinimum surface area is 108 times the cube root of 4 square inches when x = 6 times the cube root of 2 inches.\u201d\" width=\"477\" height=\"280\" \/> Figure 9. We can use a graph to determine the dimensions of a box of given the volume and the minimum surface area.[\/caption]\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Minimizing Surface Area.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/_lNDHOV2FlQ?controls=0&amp;start=1356&amp;end=1567&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266835\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266835\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.7AppliedOptimizationProblems1356to1567_transcript.txt\" target=\"_blank\" rel=\"noopener\">view the transcript for this segmented clip of \"4.7 Applied Optimization Problems\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165042713675\">Consider the same open-top box, which is to have volume [latex]216 \\, \\text{in}^3[\/latex]. Suppose the cost of the material for the base is [latex]$0.20 \/ \\text{in}^2[\/latex] and the cost of the material for the sides is [latex]$0.30 \/ \\text{in}^2[\/latex] and we are trying to minimize the cost of this box. Write the cost as a function of the side lengths of the base. (Let [latex]x[\/latex] be the side length of the base and [latex]y[\/latex] be the height of the box.)<\/p>\r\n[reveal-answer q=\"2584409\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"2584409\"]\r\n<p id=\"fs-id1165043257620\">If the cost of one of the sides is [latex]$0.30 \/ \\text{in}^2[\/latex], the cost of that side is [latex]0.30xy[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165042552669\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042552669\"]\r\n<p id=\"fs-id1165042552669\">[latex]c(x)=\\frac{259.2}{x}+0.2x^2[\/latex] dollars<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Set up and solve optimization problems in several applied fields<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165042514973\" class=\"bc-section section\">\n<h2>Solving Optimization Problems over a Closed, Bounded Interval<\/h2>\n<p id=\"fs-id1165043329454\">The basic idea of the<strong> optimization problems<\/strong> that follow is the same. We have a particular quantity that we are interested in maximizing or minimizing. However, we also have some auxiliary condition that needs to be satisfied. For instance, in the example below, we are interested in maximizing the area of a rectangular garden. Certainly, if we keep making the side lengths of the garden larger, the area will continue to become larger. However, what if we have some restriction on how much fencing we can use for the perimeter? In this case, we cannot make the garden as large as we like. Let\u2019s look at how we can maximize the area of a rectangle subject to some constraint on the perimeter.<\/p>\n<div class=\"textbook exercises\">\n<h3>Example: Maximizing the Area of a Garden<\/h3>\n<p id=\"fs-id1165042623548\">A rectangular garden is to be constructed using a rock wall as one side of the garden and wire fencing for the other three sides (Figure 1). Given 100 ft of wire fencing, determine the dimensions that would create a garden of maximum area. What is the maximum area?<\/p>\n<div style=\"width: 472px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211219\/CNX_Calc_Figure_04_07_001.jpg\" alt=\"A drawing of a garden has x and y written on the vertical and horizontal sides, respectively. There is a rock wall running along the entire bottom horizontal length of the drawing.\" width=\"462\" height=\"378\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. We want to determine the measurements [latex]x[\/latex] and [latex]y[\/latex] that will create a garden with a maximum area using 100 ft of fencing.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043262577\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043262577\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043262577\">Let [latex]x[\/latex] denote the length of the side of the garden perpendicular to the rock wall and [latex]y[\/latex] denote the length of the side parallel to the rock wall. Then the area of the garden is<\/p>\n<div id=\"fs-id1165042515463\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A=x\u00b7y[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043222797\">We want to find the maximum possible area subject to the constraint that the total fencing is [latex]100[\/latex] ft. From Figure 1, the total amount of fencing used will be [latex]2x+y[\/latex]. Therefore, the constraint equation is<\/p>\n<div id=\"fs-id1165043249986\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]2x+y=100[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043272012\">Solving this equation for [latex]y[\/latex], we have [latex]y=100-2x[\/latex]. Thus, we can write the area as<\/p>\n<div id=\"fs-id1165043281405\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A(x)=x \\cdot (100-2x)=100x-2x^2[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043195368\">Before trying to maximize the area function [latex]A(x)=100x-2x^2[\/latex], we need to determine the domain under consideration. To construct a rectangular garden, we certainly need the lengths of both sides to be positive. Therefore, we need [latex]x>0[\/latex] and [latex]y>0[\/latex]. Since [latex]y=100-2x[\/latex], if [latex]y>0[\/latex], then [latex]x<50[\/latex]. Therefore, we are trying to determine the maximum value of [latex]A(x)[\/latex] for [latex]x[\/latex] over the open interval [latex](0,50)[\/latex]. We do not know that a function necessarily has a maximum value over an open interval. However, we do know that a continuous function has an absolute maximum (and absolute minimum) over a closed interval. Therefore, let\u2019s consider the function [latex]A(x)=100x-2x^2[\/latex] over the closed interval [latex][0,50][\/latex]. If the maximum value occurs at an interior point, then we have found the value [latex]x[\/latex] in the open interval [latex](0,50)[\/latex] that maximizes the area of the garden. Therefore, we consider the following problem:<\/p>\n<p id=\"fs-id1165043094646\" style=\"text-align: center;\">Maximize [latex]A(x)=100x-2x^2[\/latex] over the interval [latex][0,50][\/latex]<\/p>\n<p id=\"fs-id1165042989293\">As mentioned earlier, since [latex]A[\/latex] is a continuous function on a closed, bounded interval, by the extreme value theorem, it has a maximum and a minimum. These extreme values occur either at endpoints or critical points. At the endpoints, [latex]A(x)=0[\/latex]. Since the area is positive for all [latex]x[\/latex] in the open interval [latex](0,50)[\/latex], the maximum must occur at a critical point. Differentiating the function [latex]A(x)[\/latex], we obtain<\/p>\n<div id=\"fs-id1165043380052\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A^{\\prime}(x)=100-4x[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043178466\">Therefore, the only critical point is [latex]x=25[\/latex] (Figure 2). We conclude that the maximum area must occur when [latex]x=25[\/latex]. Then we have [latex]y=100-2x=100-2(25)=50[\/latex]. To maximize the area of the garden, let [latex]x=25[\/latex] ft and [latex]y=50[\/latex] ft. The area of this garden is [latex]1250 \\, \\text{ft}^2[\/latex].<\/p>\n<div style=\"width: 544px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211223\/CNX_Calc_Figure_04_07_002.jpg\" alt=\"The function A(x) = 100x \u2013 2x is graphed. At its maximum there is an intersection of two dashed lines and text that reads \u201cMaximum area is 1250 square feet when x = 25 feet.\u201d\" width=\"534\" height=\"347\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. To maximize the area of the garden, we need to find the maximum value of the function [latex]A(x)=100x-2x^2[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Maximizing the Area of a Garden.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/_lNDHOV2FlQ?controls=0&amp;start=15&amp;end=180&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.7AppliedOptimizationProblems15to180_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.7 Applied Optimization Problems&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1165042474337\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165043035196\">Determine the maximum area if we want to make the same rectangular garden as in Figure 2, but we have 200 ft of fencing.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q355002\">Hint<\/span><\/p>\n<div id=\"q355002\" class=\"hidden-answer\" style=\"display: none\">\n<p>We need to maximize the function [latex]A(x)=200x-2x^2[\/latex] over the interval [latex][0,100][\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043102115\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043102115\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043102115\">The maximum area is [latex]5000 \\, \\text{ft}^2[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165042376423\">Now let\u2019s look at a general strategy for solving optimization problems similar to these above.<\/p>\n<div id=\"fs-id1165042916977\" class=\"textbox examples\">\n<h3>Problem-Solving Strategy: Solving Optimization Problems<\/h3>\n<ol id=\"fs-id1165043013367\">\n<li>Introduce all variables. If applicable, draw a figure and label all variables.<\/li>\n<li>Determine which quantity is to be maximized or minimized, and for what range of values of the other variables (if this can be determined at this time).<\/li>\n<li>Write a formula for the quantity to be maximized or minimized in terms of the variables. This formula may involve more than one variable.<\/li>\n<li>Write any equations relating the independent variables in the formula from step 3. Use these equations to write the quantity to be maximized or minimized as a function of one variable.<\/li>\n<li>Identify the domain of consideration for the function in step 4 based on the physical problem to be solved.<\/li>\n<li>Locate the maximum or minimum value of the function from step 4. This step typically involves looking for critical points and evaluating a function at endpoints.<\/li>\n<\/ol>\n<\/div>\n<p id=\"fs-id1165043029463\">Now let\u2019s apply this strategy to maximize the volume of an open-top box given a constraint on the amount of material to be used.<\/p>\n<div id=\"fs-id1165043024669\" class=\"textbook exercises\">\n<h3>example: Maximizing the Volume of a Box<\/h3>\n<p id=\"fs-id1165042638735\">An open-top box is to be made from a 24 in. by 36 in. piece of cardboard by removing a square from each corner of the box and folding up the flaps on each side. What size square should be cut out of each corner to get a box with the maximum volume?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043092635\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043092635\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043092635\">Step 1: Let [latex]x[\/latex] be the side length of the square to be removed from each corner (Figure 3). Then, the remaining four flaps can be folded up to form an open-top box. Let [latex]V[\/latex] be the volume of the resulting box.<\/p>\n<div style=\"width: 609px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211226\/CNX_Calc_Figure_04_07_003.jpg\" alt=\"There are two figures for this figure. The first one is a rectangle with sides 24 in and 36 in, with each corner having a square of side length x taken out of it. In the second picture, there is a box with side lengths x in, 24 \u2013 2x in, and 36 \u2013 2x in.\" width=\"599\" height=\"333\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. A square with side length [latex]x[\/latex] inches is removed from each corner of the piece of cardboard. The remaining flaps are folded to form an open-top box.<\/p>\n<\/div>\n<p id=\"fs-id1165042330556\">Step 2: We are trying to maximize the volume of a box. Therefore, the problem is to maximize [latex]V[\/latex].<\/p>\n<p id=\"fs-id1165043251052\">Step 3: As mentioned in step 2, are trying to maximize the volume of a box. The volume of a box is [latex]V=L \\cdot W \\cdot H[\/latex], where [latex]L, \\, W[\/latex], and [latex]H[\/latex] are the length, width, and height, respectively.<\/p>\n<p>Step 4: From Figure 3, we see that the height of the box is [latex]x[\/latex] inches, the length is [latex]36-2x[\/latex] inches, and the width is [latex]24-2x[\/latex] inches. Therefore, the volume of the box is<\/p>\n<div id=\"fs-id1165042645725\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V(x)=(36-2x)(24-2x)x=4x^3-120x^2+864x[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043115396\">Step 5: To determine the domain of consideration, let\u2019s examine Figure 3. Certainly, we need [latex]x>0[\/latex]. Furthermore, the side length of the square cannot be greater than or equal to half the length of the shorter side, 24 in.; otherwise, one of the flaps would be completely cut off. Therefore, we are trying to determine whether there is a maximum volume of the box for [latex]x[\/latex] over the open interval [latex](0,12)[\/latex]. Since [latex]V[\/latex] is a continuous function over the closed interval [latex][0,12][\/latex], we know [latex]V[\/latex] will have an absolute maximum over the closed interval. Therefore, we consider [latex]V[\/latex] over the closed interval [latex][0,12][\/latex] and check whether the absolute maximum occurs at an interior point.<\/p>\n<p id=\"fs-id1165043017956\">Step 6: Since [latex]V(x)[\/latex] is a continuous function over the closed, bounded interval [latex][0,12][\/latex], [latex]V[\/latex] must have an absolute maximum (and an absolute minimum). Since [latex]V(x)=0[\/latex] at the endpoints and [latex]V(x)>0[\/latex] for [latex]0<x<12[\/latex], the maximum must occur at a critical point. The derivative is<\/p>\n<div id=\"fs-id1165043333862\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V^{\\prime}(x)=12x^2-240x+864[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042608183\">To find the critical points, we need to solve the equation<\/p>\n<div id=\"fs-id1165042946835\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]12x^2-240x+864=0[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043209010\">Dividing both sides of this equation by 12, the problem simplifies to solving the equation<\/p>\n<div id=\"fs-id1165042982075\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x^2-20x+72=0[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043060298\">Using the quadratic formula, we find that the critical points are<\/p>\n<div id=\"fs-id1165042367652\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x=\\frac{20 \\pm \\sqrt{(-20)^2-4(1)(72)}}{2}=\\frac{20 \\pm \\sqrt{112}}{2}=\\frac{20 \\pm 4\\sqrt{7}}{2}=10 \\pm 2\\sqrt{7}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043263257\">Since [latex]10+2\\sqrt{7}[\/latex] is not in the domain of consideration, the only critical point we need to consider is [latex]10-2\\sqrt{7}[\/latex]. Therefore, the volume is maximized if we let [latex]x=10-2\\sqrt{7}[\/latex] in. The maximum volume is [latex]V(10-2\\sqrt{7})=640+448\\sqrt{7}\\approx 1825 \\, \\text{in}^3[\/latex] as shown in the following graph.<\/p>\n<div style=\"width: 729px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211230\/CNX_Calc_Figure_04_07_004.jpg\" alt=\"The function V(x) = 4x3 \u2013 120x2 + 864x is graphed. At its maximum there is an intersection of two dashed lines and text that reads \u201cMaximum volume is approximately 1825 cubic inches when x \u2248 4.7 inches.\u201d\" width=\"719\" height=\"388\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. Maximizing the volume of the box leads to finding the maximum value of a cubic polynomial.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043019959\" class=\"textbox tryit\">\n<h3>Interactive<\/h3>\n<p><a href=\"https:\/\/www.khanacademy.org\/math\/old-differential-calculus\/derivative-applications-dc\/optimization-dc\/v\/optimizing-box-volume-graphically\" target=\"_blank\" rel=\"noopener\">Watch a video about optimizing the volume of a box.<\/a><\/p>\n<\/div>\n<div id=\"fs-id1165042992008\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165043108738\">Suppose the dimensions of the cardboard in the previous example are 20 in. by 30 in. Let [latex]x[\/latex] be the side length of each square and write the volume of the open-top box as a function of [latex]x[\/latex]. Determine the domain of consideration for [latex]x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q8044661\">Hint<\/span><\/p>\n<div id=\"q8044661\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043080382\">The volume of the box is [latex]L \\cdot W \\cdot H[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043177823\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043177823\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043177823\">[latex]V(x)=x(20-2x)(30-2x)[\/latex]. The domain is [latex][0,10][\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042350080\" class=\"textbook exercises\">\n<h3>Example: Minimizing Travel Time<\/h3>\n<p id=\"fs-id1165043329514\">An island is [latex]2[\/latex] mi due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that is [latex]6[\/latex] mi west of that point. The visitor is planning to go from the cabin to the island. Suppose the visitor runs at a rate of [latex]8[\/latex] mph and swims at a rate of [latex]3[\/latex] mph. How far should the visitor run before swimming to minimize the time it takes to reach the island?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043276181\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043276181\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043276181\">Step 1: Let [latex]x[\/latex] be the distance running and let [latex]y[\/latex] be the distance swimming (Figure 5). Let [latex]T[\/latex] be the time it takes to get from the cabin to the island.<\/p>\n<div style=\"width: 604px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211233\/CNX_Calc_Figure_04_07_011.jpg\" alt=\"The cabin is x miles from the shore. From that point on the shore, the island is y miles away. If you were to continue the line from the cabin to the shore (the x miles one) and if you were to draw a line from the island parallel to the shore, then the lines would extend 2 miles from the island and 6 miles from the cabin before intersecting.\" width=\"594\" height=\"352\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5. How can we choose [latex]x[\/latex] and [latex]y[\/latex] to minimize the travel time from the cabin to the island?<\/p>\n<\/div>\n<p id=\"fs-id1165042330905\">Step 2: The problem is to minimize [latex]T[\/latex].<\/p>\n<p id=\"fs-id1165043346985\">Step 3: To find the time spent traveling from the cabin to the island, add the time spent running and the time spent swimming. Since Distance [latex]=[\/latex] Rate [latex]\\times[\/latex] Time [latex](D=R \\times T)[\/latex], the time spent running is<\/p>\n<div id=\"fs-id1165043431452\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T_{\\text{running}}=\\frac{D_{\\text{running}}}{R_{\\text{running}}}=\\frac{x}{8}[\/latex],<\/div>\n<p>and the time spent swimming is<\/p>\n<div id=\"fs-id1165042369650\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T_{\\text{swimming}}=\\frac{D_{\\text{swimming}}}{R_{\\text{swimming}}}=\\frac{y}{3}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043262597\">Therefore, the total time spent traveling is<\/p>\n<div id=\"fs-id1165043433402\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T=\\frac{x}{8}+\\frac{y}{3}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042517820\">Step 4: From Figure 5, the line segment of [latex]y[\/latex] miles forms the hypotenuse of a right triangle with legs of length [latex]2[\/latex] mi and [latex]6-x[\/latex] mi. Therefore, by the Pythagorean theorem, [latex]2^2+(6-x)^2=y^2[\/latex], and we obtain [latex]y=\\sqrt{(6-x)^2+4}[\/latex]. Thus, the total time spent traveling is given by the function<\/p>\n<div id=\"fs-id1165043430018\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T(x)=\\frac{x}{8}+\\frac{\\sqrt{(6-x)^2+4}}{3}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042328992\">Step 5: From Figure 5, we see that [latex]0\\le x\\le 6[\/latex]. Therefore, [latex][0,6][\/latex] is the domain of consideration.<\/p>\n<p id=\"fs-id1165043066617\">Step 6: Since [latex]T(x)[\/latex] is a continuous function over a closed, bounded interval, it has a maximum and a minimum. Let\u2019s begin by looking for any critical points of [latex]T[\/latex] over the interval [latex][0,6][\/latex]. The derivative is<\/p>\n<div id=\"fs-id1165043353795\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T^{\\prime}(x)=\\frac{1}{8}-\\frac{1}{2}\\frac{[(6-x)^2+4]^{-1\/2}}{3} \\cdot 2(6-x)=\\frac{1}{8}-\\frac{(6-x)}{3\\sqrt{(6-x)^2+4}}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042516379\">If [latex]T^{\\prime}(x)=0[\/latex], then<\/p>\n<div id=\"fs-id1165042705986\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{8}=\\frac{6-x}{3\\sqrt{(6-x)^2+4}}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043107403\">Therefore,<\/p>\n<div id=\"fs-id1165042343282\" class=\"equation\" style=\"text-align: center;\">[latex]3\\sqrt{(6-x)^2+4}=8(6-x)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043284394\">Squaring both sides of this equation, we see that if [latex]x[\/latex] satisfies this equation, then [latex]x[\/latex] must satisfy<\/p>\n<div id=\"fs-id1165043103774\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]9[(6-x)^2+4]=64(6-x)^2[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042923339\">which implies<\/p>\n<div id=\"fs-id1165043285288\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]55(6-x)^2=36[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042711030\">We conclude that if [latex]x[\/latex] is a critical point, then [latex]x[\/latex] satisfies<\/p>\n<div id=\"fs-id1165042923510\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](x-6)^2=\\frac{36}{55}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043396085\">Therefore, the possibilities for critical points are<\/p>\n<div id=\"fs-id1165042350053\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x=6 \\pm \\frac{6}{\\sqrt{55}}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042616899\">Since [latex]x=6+\\frac{6}{\\sqrt{55}}[\/latex] is not in the domain, it is not a possibility for a critical point. On the other hand, [latex]x=6-\\frac{6}{\\sqrt{55}}[\/latex] is in the domain. Since we squared both sides of the equation to arrive at the possible critical points, it remains to verify that it is satisfied by [latex]x=6-\\frac{6}{\\sqrt{55}}[\/latex]. Since [latex]x=6-\\frac{6}{\\sqrt{55}}[\/latex] does satisfy that equation, we conclude that [latex]x=6-\\frac{6}{\\sqrt{55}}[\/latex] is a critical point, and it is the only one. To justify that the time is minimized for this value of [latex]x[\/latex], we just need to check the values of [latex]T(x)[\/latex] at the endpoints [latex]x=0[\/latex] and [latex]x=6[\/latex], and compare them with the value of [latex]T(x)[\/latex] at the critical point [latex]x=6-\\frac{6}{\\sqrt{55}}[\/latex]. We find that [latex]T(0)\\approx 2.108[\/latex] h and [latex]T(6)\\approx 1.417[\/latex] h, whereas [latex]T(6-\\frac{6}{\\sqrt{55}})\\approx 1.368[\/latex] h. Therefore, we conclude that [latex]T[\/latex] has a local minimum at [latex]x\\approx 5.19[\/latex] mi.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043107415\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165043348922\">Suppose the island is 1 mi from shore, and the distance from the cabin to the point on the shore closest to the island is [latex]15[\/latex] mi. Suppose a visitor swims at the rate of [latex]2.5[\/latex] mph and runs at a rate of [latex]6[\/latex] mph. Let [latex]x[\/latex] denote the distance the visitor will run before swimming, and find a function for the time it takes the visitor to get from the cabin to the island.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q212787\">Hint<\/span><\/p>\n<div id=\"q212787\" class=\"hidden-answer\" style=\"display: none\">\n<p>The time [latex]T=T_{\\text{running}}+T_{\\text{swimming}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042369705\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042369705\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042369705\">[latex]T(x)=\\frac{x}{6}+\\frac{\\sqrt{(15-x)^2+1}}{2.5}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165042556259\">In business, companies are interested in <span class=\"no-emphasis\">maximizing revenue<\/span>. In the following example, we consider a scenario in which a company has collected data on how many cars it is able to lease, depending on the price it charges its customers to rent a car. Let\u2019s use these data to determine the price the company should charge to maximize the amount of money it brings in.<\/p>\n<div id=\"fs-id1165043353574\" class=\"textbook exercises\">\n<h3>Example: Maximizing Revenue<\/h3>\n<p id=\"fs-id1165042515832\">Owners of a car rental company have determined that if they charge customers [latex]p[\/latex] dollars per day to rent a car, where [latex]50\\le p\\le 200[\/latex], the number of cars [latex]n[\/latex] they rent per day can be modeled by the linear function [latex]n(p)=1000-5p[\/latex]. If they charge [latex]$50[\/latex] per day or less, they will rent all their cars. If they charge [latex]$200[\/latex] per day or more, they will not rent any cars. Assuming the owners plan to charge customers between [latex]$50[\/latex] per day and [latex]$200[\/latex] per day to rent a car, how much should they charge to maximize their revenue?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043423441\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043423441\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043423441\">Step 1: Let [latex]p[\/latex] be the price charged per car per day and let [latex]n[\/latex] be the number of cars rented per day. Let [latex]R[\/latex] be the revenue per day.<\/p>\n<p id=\"fs-id1165043431290\">Step 2: The problem is to maximize [latex]R[\/latex].<\/p>\n<p id=\"fs-id1165043087644\">Step 3: The revenue (per day) is equal to the number of cars rented per day times the price charged per car per day\u2014that is, [latex]R=n \\times p[\/latex].<\/p>\n<p id=\"fs-id1165042514435\">Step 4: Since the number of cars rented per day is modeled by the linear function [latex]n(p)=1000-5p[\/latex], the revenue [latex]R[\/latex] can be represented by the function<\/p>\n<div id=\"fs-id1165043256835\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]R(p)=n \\times p=(1000-5p)p=-5p^2+1000p[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042406610\">Step 5: Since the owners plan to charge between [latex]$50[\/latex] per car per day and [latex]$200[\/latex] per car per day, the problem is to find the maximum revenue [latex]R(p)[\/latex] for [latex]p[\/latex] in the closed interval [latex][50,200][\/latex].<\/p>\n<p id=\"fs-id1165043173962\">Step 6: Since [latex]R[\/latex] is a continuous function over the closed, bounded interval [latex][50,200][\/latex], it has an absolute maximum (and an absolute minimum) in that interval. To find the maximum value, look for critical points. The derivative is [latex]R^{\\prime}(p)=-10p+1000[\/latex]. Therefore, the critical point is [latex]p=100[\/latex] When [latex]p=100[\/latex], [latex]R(100)=$50,000[\/latex]. When [latex]p=50[\/latex], [latex]R(p)=$37,500[\/latex]. When [latex]p=200[\/latex], [latex]R(p)=$0[\/latex]. Therefore, the absolute maximum occurs at [latex]p=$100[\/latex]. The car rental company should charge [latex]$100[\/latex] per day per car to maximize revenue as shown in the following figure.<\/p>\n<div style=\"width: 505px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211236\/CNX_Calc_Figure_04_07_010.jpg\" alt=\"The function R(p) is graphed. At its maximum there is an intersection of two dashed lines and text that reads \u201cMaximum revenue is 50,000 per day when the price charged per car is 100 per day.\u201d\" width=\"495\" height=\"270\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. To maximize revenue, a car rental company has to balance the price of a rental against the number of cars people will rent at that price.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Maximizing Revenue.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/_lNDHOV2FlQ?controls=0&amp;start=913&amp;end=1024&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.7AppliedOptimizationProblems913to1024_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.7 Applied Optimization Problems&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1165043257501\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165042376530\">A car rental company charges its customers [latex]p[\/latex] dollars per day, where [latex]60\\le p\\le 150[\/latex]. It has found that the number of cars rented per day can be modeled by the linear function [latex]n(p)=750-5p[\/latex]. How much should the company charge each customer to maximize revenue?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q13435467\">Hint<\/span><\/p>\n<div id=\"q13435467\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042321515\">[latex]R(p)=n \\times p[\/latex], where [latex]n[\/latex] is the number of cars rented and [latex]p[\/latex] is the price charged per car.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042960136\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042960136\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042960136\">The company should charge [latex]$75[\/latex] per car per day.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043350161\" class=\"textbook exercises\">\n<h3>Example: Maximizing the Area of an Inscribed Rectangle<\/h3>\n<p id=\"fs-id1165042456921\">A rectangle is to be inscribed in the ellipse<\/p>\n<div id=\"fs-id1165042456924\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{x^2}{4}+y^2=1[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042645676\">What should the dimensions of the rectangle be to maximize its area? What is the maximum area?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042519826\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042519826\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042519826\">Step 1: For a rectangle to be inscribed in the ellipse, the sides of the rectangle must be parallel to the axes. Let [latex]L[\/latex] be the length of the rectangle and [latex]W[\/latex] be its width. Let [latex]A[\/latex] be the area of the rectangle.<\/p>\n<div style=\"width: 585px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211239\/CNX_Calc_Figure_04_07_007.jpg\" alt=\"The ellipse x2\/4 + y2 = 1 is drawn with its x intercepts being \u00b12 and its y intercepts being \u00b11. There is a rectangle inscribed in the ellipse with length L (in the x-direction) and width W.\" width=\"575\" height=\"350\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 7. We want to maximize the area of a rectangle inscribed in an ellipse.<\/p>\n<\/div>\n<p id=\"fs-id1165042371013\">Step 2: The problem is to maximize [latex]A[\/latex].<\/p>\n<p id=\"fs-id1165042377481\">Step 3: The area of the rectangle is [latex]A=L \\times W[\/latex].<\/p>\n<p id=\"fs-id1165043257695\">Step 4: Let [latex](x,y)[\/latex] be the corner of the rectangle that lies in the first quadrant, as shown in Figure 7. We can write length [latex]L=2x[\/latex] and width [latex]W=2y[\/latex]. Since [latex]\\frac{x^2}{4}+y^2=1[\/latex] and [latex]y>0[\/latex], we have [latex]y=\\sqrt{1 - \\frac{x^2}{4}}[\/latex]. Therefore, the area is<\/p>\n<div id=\"fs-id1165042320390\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A=L \\times W=(2x)(2y)=4x\\sqrt{1 - \\frac{x^2}{4}}=2x\\sqrt{4-x^2}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043380076\">Step 5: From Figure 7, we see that to inscribe a rectangle in the ellipse, the [latex]x[\/latex]-coordinate of the corner in the first quadrant must satisfy [latex]0<x<2[\/latex]. Therefore, the problem reduces to looking for the maximum value of [latex]A(x)[\/latex] over the open interval [latex](0,2)[\/latex]. Since [latex]A(x)[\/latex] will have an absolute maximum (and absolute minimum) over the closed interval [latex][0,2][\/latex], we consider [latex]A(x)=2x\\sqrt{4-x^2}[\/latex] over the interval [latex][0,2][\/latex]. If the absolute maximum occurs at an interior point, then we have found an absolute maximum in the open interval.<\/p>\n<p id=\"fs-id1165042331466\">Step 6: As mentioned earlier, [latex]A(x)[\/latex] is a continuous function over the closed, bounded interval [latex][0,2][\/latex]. Therefore, it has an absolute maximum (and absolute minimum). At the endpoints [latex]x=0[\/latex] and [latex]x=2[\/latex], [latex]A(x)=0[\/latex]. For [latex]0<x<2[\/latex], [latex]A(x)>0[\/latex]. Therefore, the maximum must occur at a critical point. Taking the derivative of [latex]A(x)[\/latex], we obtain<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} A^{\\prime}(x) & =2\\sqrt{4-x^2}+2x \\cdot \\frac{1}{2\\sqrt{4-x^2}}(-2x) \\\\ & =2\\sqrt{4-x^2}-\\frac{2x^2}{\\sqrt{4-x^2}} \\\\ & =\\frac{8-4x^2}{\\sqrt{4-x^2}} \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043379942\">To find critical points, we need to find where [latex]A^{\\prime}(x)=0[\/latex]. We can see that if [latex]x[\/latex] is a solution of<\/p>\n<div id=\"fs-id1165043354675\" class=\"equation\" style=\"text-align: center;\">[latex]\\frac{8-4x^2}{\\sqrt{4-x^2}}=0[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043183716\">then [latex]x[\/latex] must satisfy<\/p>\n<div id=\"fs-id1165043281800\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]8-4x^2=0[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042515040\">Therefore, [latex]x^2=2[\/latex]. Thus, [latex]x=\\pm \\sqrt{2}[\/latex] are the possible solutions of[latex]A^{\\prime}(x)[\/latex]. Since we are considering [latex]x[\/latex] over the interval [latex][0,2][\/latex], [latex]x=\\sqrt{2}[\/latex] is a possibility for a critical point, but [latex]x=\u2212\\sqrt{2}[\/latex] is not. Therefore, we check whether [latex]\\sqrt{2}[\/latex] is a solution of [latex]A^{\\prime}(x)[\/latex]. Since [latex]x=\\sqrt{2}[\/latex] is a solution of[latex]A^{\\prime}(x)[\/latex], we conclude that [latex]\\sqrt{2}[\/latex] is the only critical point of [latex]A(x)[\/latex] in the interval [latex][0,2][\/latex]. Therefore, [latex]A(x)[\/latex] must have an absolute maximum at the critical point [latex]x=\\sqrt{2}[\/latex]. To determine the dimensions of the rectangle, we need to find the length [latex]L[\/latex] and the width [latex]W[\/latex]. If [latex]x=\\sqrt{2}[\/latex] then<\/p>\n<div id=\"fs-id1165042708515\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\sqrt{1-\\frac{(\\sqrt{2})^2}{4}}=\\sqrt{1-\\frac{1}{2}}=\\frac{1}{\\sqrt{2}}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043376531\">Therefore, the dimensions of the rectangle are [latex]L=2x=2\\sqrt{2}[\/latex] and [latex]W=2y=\\frac{2}{\\sqrt{2}}=\\sqrt{2}[\/latex]. The area of this rectangle is [latex]A=L \\times W=(2\\sqrt{2})(\\sqrt{2})=4[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043390986\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165043199268\">Modify the area function [latex]A[\/latex] if the rectangle is to be inscribed in the unit circle [latex]x^2+y^2=1[\/latex]. What is the domain of consideration?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q70845661\">Hint<\/span><\/p>\n<div id=\"q70845661\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042318676\">If [latex](x,y)[\/latex] is the vertex of the square that lies in the first quadrant, then the area of the square is [latex]A=(2x)(2y)=4xy[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043276112\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043276112\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043276112\">[latex]A(x)=4x\\sqrt{1-x^2}[\/latex]. The domain of consideration is [latex][0,1][\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm54694\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=54694&theme=oea&iframe_resize_id=ohm54694&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2 data-type=\"title\">Solving Optimization Problems when the Interval Is Not Closed or Is Unbounded<\/h2>\n<p id=\"fs-id1165042318512\">In the previous examples, we considered functions on closed, bounded domains. Consequently, by the extreme value theorem, we were guaranteed that the functions had absolute extrema. Let\u2019s now consider functions for which the domain is neither closed nor bounded.<\/p>\n<p id=\"fs-id1165043173977\">Many functions still have at least one absolute extrema, even if the domain is not closed or the domain is unbounded. For example, the function [latex]f(x)=x^2+4[\/latex] over [latex](\u2212\\infty ,\\infty )[\/latex] has an absolute minimum of 4 at [latex]x=0[\/latex]. Therefore, we can still consider functions over unbounded domains or open intervals and determine whether they have any absolute extrema. In the next example, we try to minimize a function over an unbounded domain. We will see that, although the domain of consideration is [latex](0,\\infty )[\/latex], the function has an absolute minimum.<\/p>\n<p id=\"fs-id1165042610536\">In the following example, we look at constructing a box of least surface area with a prescribed volume. It is not difficult to show that for a closed-top box, by symmetry, among all boxes with a specified volume, a cube will have the smallest surface area. Consequently, we consider the modified problem of determining which open-topped box with a specified volume has the smallest surface area.<\/p>\n<div id=\"fs-id1165042327673\" class=\"textbook exercises\">\n<h3>Example: Minimizing Surface Area<\/h3>\n<p id=\"fs-id1165043430067\">A rectangular box with a square base, an open top, and a volume of [latex]216 \\, \\text{in}^3[\/latex]\u00a0is to be constructed. What should the dimensions of the box be to minimize the surface area of the box? What is the minimum surface area?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042373482\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042373482\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042373482\">Step 1: Draw a rectangular box and introduce the variable [latex]x[\/latex] to represent the length of each side of the square base; let [latex]y[\/latex] represent the height of the box. Let [latex]S[\/latex] denote the surface area of the open-top box.<\/p>\n<div style=\"width: 291px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211241\/CNX_Calc_Figure_04_07_008.jpg\" alt=\"A box with square base is shown. The base has side length x, and the height is y.\" width=\"281\" height=\"264\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 8. We want to minimize the surface area of a square-based box with a given volume.<\/p>\n<\/div>\n<p id=\"fs-id1165042355751\">Step 2: We need to minimize the surface area. Therefore, we need to minimize [latex]S[\/latex].<\/p>\n<p id=\"fs-id1165043430591\">Step 3: Since the box has an open top, we need only determine the area of the four vertical sides and the base. The area of each of the four vertical sides is [latex]x \\cdot y[\/latex]. The area of the base is [latex]x^2[\/latex]. Therefore, the surface area of the box is<\/p>\n<div id=\"fs-id1165042327660\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]S=4xy+x^2[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042322398\">Step 4: Since the volume of this box is [latex]x^2 y[\/latex] and the volume is given as [latex]216 \\, \\text{in}^3[\/latex], the constraint equation is<\/p>\n<div id=\"fs-id1165042948321\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x^2 y=216[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042418854\">Solving the constraint equation for [latex]y[\/latex], we have [latex]y=\\frac{216}{x^2}[\/latex]. Therefore, we can write the surface area as a function of [latex]x[\/latex] only:<\/p>\n<div id=\"fs-id1165042375810\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]S(x)=4x(\\frac{216}{x^2})+x^2[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042318390\">Therefore, [latex]S(x)=\\frac{864}{x}+x^2[\/latex].<\/p>\n<p id=\"fs-id1165043393971\">Step 5: Since we are requiring that [latex]x^2 y=216[\/latex], we cannot have [latex]x=0[\/latex]. Therefore, we need [latex]x>0[\/latex]. On the other hand, [latex]x[\/latex] is allowed to have any positive value. Note that as [latex]x[\/latex] becomes large, the height of the box [latex]y[\/latex] becomes correspondingly small so that [latex]x^2 y=216[\/latex]. Similarly, as [latex]x[\/latex] becomes small, the height of the box becomes correspondingly large. We conclude that the domain is the open, unbounded interval [latex](0,\\infty )[\/latex]. Note that, unlike the previous examples, we cannot reduce our problem to looking for an absolute maximum or absolute minimum over a closed, bounded interval. However, in the next step, we discover why this function must have an absolute minimum over the interval [latex](0,\\infty )[\/latex].<\/p>\n<p id=\"fs-id1165042638460\">Step 6: Note that as [latex]x\\to 0^+[\/latex], [latex]S(x)\\to \\infty[\/latex]. Also, as [latex]x\\to \\infty[\/latex], [latex]S(x)\\to \\infty[\/latex]. Since [latex]S[\/latex] is a continuous function that approaches infinity at the ends, it must have an absolute minimum at some [latex]x\\in (0,\\infty )[\/latex]. This minimum must occur at a critical point of [latex]S[\/latex]. The derivative is<\/p>\n<div id=\"fs-id1165043430791\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]S^{\\prime}(x)=-\\frac{864}{x^2}+2x[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042355317\">Therefore, [latex]S^{\\prime}(x)=0[\/latex] when [latex]2x=\\frac{864}{x^2}[\/latex]. Solving this equation for [latex]x[\/latex], we obtain [latex]x^3=432[\/latex], so [latex]x=\\sqrt[3]{432}=6\\sqrt[3]{2}[\/latex]. Since this is the only critical point of [latex]S[\/latex], the absolute minimum must occur at [latex]x=6\\sqrt[3]{2}[\/latex] (see Figure 9). When [latex]x=6\\sqrt[3]{2}[\/latex], [latex]y=\\frac{216}{(6\\sqrt[3]{2})^2}=3\\sqrt[3]{2}[\/latex] in. Therefore, the dimensions of the box should be [latex]x=6\\sqrt[3]{2}[\/latex] in and [latex]y=3\\sqrt[3]{2}[\/latex] in. With these dimensions, the surface area is<\/p>\n<div id=\"fs-id1165043379912\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]S(6\\sqrt[3]{2})=\\frac{864}{6\\sqrt[3]{2}}+(6\\sqrt[3]{2})^2=108\\sqrt[3]{4} \\, \\text{in}^2[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<div>\n<div style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211244\/CNX_Calc_Figure_04_07_009.jpg\" alt=\"The function S(x) = 864\/x + x2 is graphed. At its minimum there is a dashed line and text that reads \u201cMinimum surface area is 108 times the cube root of 4 square inches when x = 6 times the cube root of 2 inches.\u201d\" width=\"477\" height=\"280\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 9. We can use a graph to determine the dimensions of a box of given the volume and the minimum surface area.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Minimizing Surface Area.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/_lNDHOV2FlQ?controls=0&amp;start=1356&amp;end=1567&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266835\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266835\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.7AppliedOptimizationProblems1356to1567_transcript.txt\" target=\"_blank\" rel=\"noopener\">view the transcript for this segmented clip of &#8220;4.7 Applied Optimization Problems&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165042713675\">Consider the same open-top box, which is to have volume [latex]216 \\, \\text{in}^3[\/latex]. Suppose the cost of the material for the base is [latex]$0.20 \/ \\text{in}^2[\/latex] and the cost of the material for the sides is [latex]$0.30 \/ \\text{in}^2[\/latex] and we are trying to minimize the cost of this box. Write the cost as a function of the side lengths of the base. (Let [latex]x[\/latex] be the side length of the base and [latex]y[\/latex] be the height of the box.)<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q2584409\">Hint<\/span><\/p>\n<div id=\"q2584409\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043257620\">If the cost of one of the sides is [latex]$0.30 \/ \\text{in}^2[\/latex], the cost of that side is [latex]0.30xy[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042552669\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042552669\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042552669\">[latex]c(x)=\\frac{259.2}{x}+0.2x^2[\/latex] dollars<\/p>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-410\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>4.7 Applied Optimization Problems. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":25,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"4.7 Applied Optimization Problems\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-410","chapter","type-chapter","status-publish","hentry"],"part":48,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/410","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":23,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/410\/revisions"}],"predecessor-version":[{"id":4845,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/410\/revisions\/4845"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/48"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/410\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=410"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=410"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=410"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=410"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}