{"id":3897,"date":"2021-05-20T18:29:56","date_gmt":"2021-05-20T18:29:56","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/review-for-applied-optimization-problems\/"},"modified":"2021-07-03T19:07:29","modified_gmt":"2021-07-03T19:07:29","slug":"review-for-applied-optimization-problems","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/review-for-applied-optimization-problems\/","title":{"raw":"Skills Review for Applied Optimization Problems","rendered":"Skills Review for Applied Optimization Problems"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Calculate the volume of rectangular solid<\/li>\r\n \t<li>Use the Pythagorean theorem to find the unknown side length of a right triangle<\/li>\r\n \t<li>Use the distance, rate, and time formula<\/li>\r\n \t<li>Write an equation in one variable to solve problems with multiple unknowns<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Applied Optimization Problems section, we will use formulas to model real-life scenarios. Calculus will then be used to either maximize or minimize the given scenario. For example, we can find the maximum area we can enclose with a given amount of fence. To review some of the formulas needed for the Applied Optimization Problems section, see Skills Review for Related Rates. Specifically, pay close attention to the outcomes listed above.\r\n<h2>One Variable for Multiple Unknowns<\/h2>\r\nWe sometimes define unknown measurements or values based on another unknown. This allows us to solve for a single variable rather than having multiple variables complicating the algebra.\r\n\r\nWhen dealing with real-world applications, there are certain expressions that we can translate directly into math. The table below lists some common verbal expressions and their equivalent mathematical expressions.\r\n<table summary=\"A table with 8 rows and 2 columns. The entries in the first row are: Verbal and Translation to math operations. The entries in the second row are: One number exceeds another by a and x, x+a. The entries in the third row are: Twice a number and 2x. The entries in the fourth row are: One number is a more than another number and x, x plus a. The entries in the fifth row are: One number is a less than twice another number and x,2 times x minus a. The entries in the sixth row are: The product of a number and a, decreased by b and a times x minus b. The entries in the seventh row are: The quotient of a number and the number plus a is three times the number and x divided by the quantity x plus a equals three times x. The entries in the eighth row are: The product of three times a number and the number decreased by b is c and three times x times the quantity x minus b equals c.\">\r\n<thead>\r\n<tr>\r\n<th>Verbal<\/th>\r\n<th>Translation to Math Operations<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>One number exceeds another by <em>a<\/em><\/td>\r\n<td>[latex]x,\\text{ }x+a[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Twice a number<\/td>\r\n<td>[latex]2x[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>One number is <em>a <\/em>more than another number<\/td>\r\n<td>[latex]x,\\text{ }x+a[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>One number is <em>a <\/em>less than twice another number<\/td>\r\n<td>[latex]x,2x-a[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>The product of a number and <em>a<\/em>, decreased by <em>b<\/em><\/td>\r\n<td>[latex]ax-b[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>The quotient of a number and the number plus <em>a <\/em>is three times the number<\/td>\r\n<td>[latex]\\Large\\frac{x}{x+a}\\normalsize =3x[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>The product of three times a number and the number decreased by <em>b <\/em>is <em>c<\/em><\/td>\r\n<td>[latex]3x\\left(x-b\\right)=c[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n&nbsp;\r\n<div class=\"textbox\">\r\n<h3>How To: Given a real-world situation, write a linear equation to model it<\/h3>\r\n<ol>\r\n \t<li>Identify known and unknown quantities.<\/li>\r\n \t<li>Assign a variable to represent the unknown quantity.<\/li>\r\n \t<li>If there is more than one unknown quantity, find a way to write the second unknown in terms of the first.<\/li>\r\n \t<li>Write an equation interpreting the words in the problem as mathematical operations.<\/li>\r\n \t<li>Solve the equation, check to be sure your answer is reasonable, and give the answer using the language and units of the original situation.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFind a linear equation to solve for the following unknown quantities: One number exceeds another number by [latex]17[\/latex] and their sum is [latex]31[\/latex]. Find the two numbers.\r\n\r\n[reveal-answer q=\"460075\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"460075\"]\r\n\r\nLet [latex]x[\/latex] equal the first number. Then, as the second number exceeds the first by 17, we can write the second number as [latex]x+17[\/latex]. The sum of the two numbers is 31. We usually interpret the word <em>is<\/em> as an equal sign.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{rl}x+\\left(x+17\\right)&amp;=31\\hfill \\\\ 2x+17&amp;=31\\hfill&amp;\\text{Simplify and solve}.\\hfill \\\\ 2x&amp;=14\\hfill \\\\ x&amp;=7\\hfill\\end{array}[\/latex]<\/div>\r\n<div><\/div>\r\n<div>The second number would then be [latex]x+17=7+17=24[\/latex]<\/div>\r\n<div><\/div>\r\n<div>The two numbers are [latex]7[\/latex] and [latex]24[\/latex].<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind a linear equation to solve for the following unknown quantities: One number is three more than twice another number. If the sum of the two numbers is [latex]36[\/latex], find the numbers.\r\n[reveal-answer q=\"930268\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"930268\"]\r\n\r\n11 and 25[\/hidden-answer]\r\n\r\n[ohm_question]142770-142775[\/ohm_question]\r\n\r\n<\/div>\r\nWe can also use a linear equation in one variable to solve a problem with two unknowns by writing an expression for one unknown in terms of the other.\r\n<div class=\"textbox exercises\">\r\n<h3>example: write a linear equation in one variable to model and solve an application<\/h3>\r\nA bag is filled with green and blue marbles. There are 77 marbles in the bag. If there are 17 more green marbles than blue marbles, find the number of green marbles and the number of blue marbles in the bag.\r\n\r\nHow many marbles of each color are in the bag?\r\n\r\n[reveal-answer q=\"470923\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"470923\"]\r\n\r\nLet [latex]G[\/latex] represent the number of green marbles in the bag and [latex]B[\/latex] represent the number of blue marbles. We have that [latex]G + B = 77[\/latex]. But we also have that there are 17 more green marbles than blue. That is, the number of green marbles is the same as the number of blue marbles plus 17. We can translate that as [latex]G = B+17[\/latex]. Since we've found a way to express the variable [latex]G[\/latex] in terms of [latex]B[\/latex], we can write one equation in one variable.\r\n\r\n[latex]B+17+B=77[\/latex], which we can solve for [latex]B[\/latex].\r\n\r\n[latex]2B=60[\/latex], so [latex]B=30[\/latex]\r\n\r\nThere are [latex]30[\/latex] Blue marbles in the bag and [latex]30+17 = 47[\/latex] is the number of green marbles.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video for another example of using a linear equation in one variable to solve an application.\r\n\r\nhttps:\/\/youtu.be\/vOA7SGQCpr8\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\nA cash register contains only five dollar and ten dollar bills. It contains twice as many five's as ten's and the total amount of money in the cash register is 560 dollars. How many ten's are in the cash register?\r\n\r\n[reveal-answer q=\"315017\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"315017\"]There are 28 ten dollar bills in the cash register.[\/hidden-answer]\r\n\r\n[ohm_question]13829[\/ohm_question]\r\n\r\n<\/div>\r\nMany applications are solved using known formulas. The problem is stated, a formula is identified, the known quantities are substituted into the formula, the equation is solved for the unknown, and the problem\u2019s question is answered. Sometimes, these problems involve two equations representing two unknowns, which can be written using one equation in one variable by expressing one unknown in terms of the other. Examples of formulas include the <strong>perimeter<\/strong> of a rectangle, [latex]P=2L+2W[\/latex]; the <strong>area<\/strong> of a rectangular region, [latex]A=LW[\/latex]; and the <strong>volume<\/strong> of a rectangular solid, [latex]V=LWH[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Perimeter Problem<\/h3>\r\nThe perimeter of a rectangular outdoor patio is [latex]54[\/latex] ft. The length is [latex]3[\/latex] ft. greater than the width. What are the dimensions of the patio?\r\n[reveal-answer q=\"815614\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"815614\"]\r\n\r\nThe perimeter formula is standard: [latex]P=2L+2W[\/latex]. We have two unknown quantities, length and width. However, we can write the length in terms of the width as [latex]L=W+3[\/latex]. Substitute the perimeter value and the expression for length into the formula. It is often helpful to make a sketch and label the sides as shown below.\r\n\r\n&nbsp;\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200341\/CNX_CAT_Figure_02_03_003.jpg\" alt=\"A rectangle with the length labeled as: L = W + 3 and the width labeled as: W.\" width=\"487\" height=\"166\" \/>\r\n\r\nNow we can solve for the width and then calculate the length.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}P=2L+2W\\hfill \\\\ 54=2\\left(W+3\\right)+2W\\hfill \\\\ 54=2W+6+2W\\hfill \\\\ 54=4W+6\\hfill \\\\ 48=4W\\hfill \\\\ 12=W\\hfill \\\\ \\left(12+3\\right)=L\\hfill \\\\ 15=L\\hfill \\end{array}[\/latex]<\/div>\r\nThe dimensions are [latex]L=15[\/latex] ft and [latex]W=12[\/latex] ft.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Evaluating a variable for an expression<\/h3>\r\nIn the example above, the length was written in terms of the width in order to substitute it for the variable [latex]L[\/latex] to be able to write one equation in one variable.\r\n\r\nThis is a powerful technique in algebra and should be practiced until it becomes familiar, so it's worth taking a closer look.\r\n\r\n[reveal-answer q=\"866900\"]more[\/reveal-answer]\r\n[hidden-answer a=\"866900\"]\r\n\r\nWe had [latex]P = 2L + 2W[\/latex], which is one equation in two variables. Given the value for [latex]P[\/latex], we don't have enough information to solve for either [latex]L[\/latex] or [latex]W[\/latex].\r\n\r\nBut we do know that [latex]L=W+3[\/latex]. We can use this to substitute for the [latex]L[\/latex] in the original equation. Be careful to retain the original statement exactly, substituting only the value for [latex]L[\/latex]. A handy way to handle the substitution is with a set of parentheses.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}P=2L+2W\\hfill \\\\ 54=2\\left(L\\right)+2W \\,\\,\\,\\,\\,\\,\\,\\, \\text{ wrap the } L \\text{ to see where to make the substitution} \\hfill \\\\ 54=2\\left(W+3\\right)+2W \\,\\,\\,\\,\\, \\text{ drop the expression for } L \\text{ into the parentheses}\\hfill \\\\ 54=2W+6+2W\\hfill \\\\ 54=4W+6\\end{array}[\/latex]<\/p>\r\nNow we have one equation in one variable. We can use the properties of equality to isolate the variable [latex]W[\/latex] on one side, with its numerical value on the other.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the dimensions of a rectangle given that the perimeter is [latex]110[\/latex] cm. and the length is 1 cm. more than twice the width.\r\n[reveal-answer q=\"447289\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"447289\"]\r\n\r\nL = 37 cm, W = 18 cm[\/hidden-answer]\r\n\r\n[ohm_question]7679[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving an Area Problem<\/h3>\r\nThe perimeter of a tablet of graph paper is 48 in<sup>2<\/sup>. The length is [latex]6[\/latex] in. more than the width. Find the area of the graph paper.\r\n[reveal-answer q=\"81698\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"81698\"]\r\n\r\nThe standard formula for area is [latex]A=LW[\/latex]; however, we will solve the problem using the perimeter formula. The reason we use the perimeter formula is because we know enough information about the perimeter that the formula will allow us to solve for one of the unknowns. As both perimeter and area use length and width as dimensions, they are often used together to solve a problem such as this one.\r\n\r\nWe know that the length is 6 in. more than the width, so we can write length as [latex]L=W+6[\/latex]. Substitute the value of the perimeter and the expression for length into the perimeter formula and find the length.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}P=2L+2W\\hfill \\\\ 48=2\\left(W+6\\right)+2W\\hfill \\\\ 48=2W+12+2W\\hfill \\\\ 48=4W+12\\hfill \\\\ 36=4W\\hfill \\\\ 9=W\\hfill \\\\ \\left(9+6\\right)=L\\hfill \\\\ 15=L\\hfill \\end{array}[\/latex]<\/div>\r\nNow, we find the area given the dimensions of [latex]L=15[\/latex] in. and [latex]W=9[\/latex] in.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}A\\hfill&amp;=LW\\hfill \\\\ A\\hfill&amp;=15\\left(9\\right)\\hfill \\\\ \\hfill&amp;=135\\text{ in}^{2}\\hfill \\end{array}[\/latex]<\/div>\r\nThe area is [latex]135[\/latex] in<sup>2<\/sup>.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nA game room has a perimeter of 70 ft. The length is five more than twice the width. How many ft<sup>2<\/sup> of new carpeting should be ordered?\r\n[reveal-answer q=\"646790\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"646790\"]\r\n\r\n250 ft<sup>2<\/sup>[\/hidden-answer]\r\n\r\n[ohm_question]1688[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Calculate the volume of rectangular solid<\/li>\n<li>Use the Pythagorean theorem to find the unknown side length of a right triangle<\/li>\n<li>Use the distance, rate, and time formula<\/li>\n<li>Write an equation in one variable to solve problems with multiple unknowns<\/li>\n<\/ul>\n<\/div>\n<p>In the Applied Optimization Problems section, we will use formulas to model real-life scenarios. Calculus will then be used to either maximize or minimize the given scenario. For example, we can find the maximum area we can enclose with a given amount of fence. To review some of the formulas needed for the Applied Optimization Problems section, see Skills Review for Related Rates. Specifically, pay close attention to the outcomes listed above.<\/p>\n<h2>One Variable for Multiple Unknowns<\/h2>\n<p>We sometimes define unknown measurements or values based on another unknown. This allows us to solve for a single variable rather than having multiple variables complicating the algebra.<\/p>\n<p>When dealing with real-world applications, there are certain expressions that we can translate directly into math. The table below lists some common verbal expressions and their equivalent mathematical expressions.<\/p>\n<table summary=\"A table with 8 rows and 2 columns. The entries in the first row are: Verbal and Translation to math operations. The entries in the second row are: One number exceeds another by a and x, x+a. The entries in the third row are: Twice a number and 2x. The entries in the fourth row are: One number is a more than another number and x, x plus a. The entries in the fifth row are: One number is a less than twice another number and x,2 times x minus a. The entries in the sixth row are: The product of a number and a, decreased by b and a times x minus b. The entries in the seventh row are: The quotient of a number and the number plus a is three times the number and x divided by the quantity x plus a equals three times x. The entries in the eighth row are: The product of three times a number and the number decreased by b is c and three times x times the quantity x minus b equals c.\">\n<thead>\n<tr>\n<th>Verbal<\/th>\n<th>Translation to Math Operations<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>One number exceeds another by <em>a<\/em><\/td>\n<td>[latex]x,\\text{ }x+a[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Twice a number<\/td>\n<td>[latex]2x[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>One number is <em>a <\/em>more than another number<\/td>\n<td>[latex]x,\\text{ }x+a[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>One number is <em>a <\/em>less than twice another number<\/td>\n<td>[latex]x,2x-a[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>The product of a number and <em>a<\/em>, decreased by <em>b<\/em><\/td>\n<td>[latex]ax-b[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>The quotient of a number and the number plus <em>a <\/em>is three times the number<\/td>\n<td>[latex]\\Large\\frac{x}{x+a}\\normalsize =3x[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>The product of three times a number and the number decreased by <em>b <\/em>is <em>c<\/em><\/td>\n<td>[latex]3x\\left(x-b\\right)=c[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a real-world situation, write a linear equation to model it<\/h3>\n<ol>\n<li>Identify known and unknown quantities.<\/li>\n<li>Assign a variable to represent the unknown quantity.<\/li>\n<li>If there is more than one unknown quantity, find a way to write the second unknown in terms of the first.<\/li>\n<li>Write an equation interpreting the words in the problem as mathematical operations.<\/li>\n<li>Solve the equation, check to be sure your answer is reasonable, and give the answer using the language and units of the original situation.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Find a linear equation to solve for the following unknown quantities: One number exceeds another number by [latex]17[\/latex] and their sum is [latex]31[\/latex]. Find the two numbers.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q460075\">Show Solution<\/span><\/p>\n<div id=\"q460075\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let [latex]x[\/latex] equal the first number. Then, as the second number exceeds the first by 17, we can write the second number as [latex]x+17[\/latex]. The sum of the two numbers is 31. We usually interpret the word <em>is<\/em> as an equal sign.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{rl}x+\\left(x+17\\right)&=31\\hfill \\\\ 2x+17&=31\\hfill&\\text{Simplify and solve}.\\hfill \\\\ 2x&=14\\hfill \\\\ x&=7\\hfill\\end{array}[\/latex]<\/div>\n<div><\/div>\n<div>The second number would then be [latex]x+17=7+17=24[\/latex]<\/div>\n<div><\/div>\n<div>The two numbers are [latex]7[\/latex] and [latex]24[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find a linear equation to solve for the following unknown quantities: One number is three more than twice another number. If the sum of the two numbers is [latex]36[\/latex], find the numbers.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q930268\">Show Solution<\/span><\/p>\n<div id=\"q930268\" class=\"hidden-answer\" style=\"display: none\">\n<p>11 and 25<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm142770\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=142770-142775&theme=oea&iframe_resize_id=ohm142770&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>We can also use a linear equation in one variable to solve a problem with two unknowns by writing an expression for one unknown in terms of the other.<\/p>\n<div class=\"textbox exercises\">\n<h3>example: write a linear equation in one variable to model and solve an application<\/h3>\n<p>A bag is filled with green and blue marbles. There are 77 marbles in the bag. If there are 17 more green marbles than blue marbles, find the number of green marbles and the number of blue marbles in the bag.<\/p>\n<p>How many marbles of each color are in the bag?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q470923\">Show Solution<\/span><\/p>\n<div id=\"q470923\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let [latex]G[\/latex] represent the number of green marbles in the bag and [latex]B[\/latex] represent the number of blue marbles. We have that [latex]G + B = 77[\/latex]. But we also have that there are 17 more green marbles than blue. That is, the number of green marbles is the same as the number of blue marbles plus 17. We can translate that as [latex]G = B+17[\/latex]. Since we&#8217;ve found a way to express the variable [latex]G[\/latex] in terms of [latex]B[\/latex], we can write one equation in one variable.<\/p>\n<p>[latex]B+17+B=77[\/latex], which we can solve for [latex]B[\/latex].<\/p>\n<p>[latex]2B=60[\/latex], so [latex]B=30[\/latex]<\/p>\n<p>There are [latex]30[\/latex] Blue marbles in the bag and [latex]30+17 = 47[\/latex] is the number of green marbles.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video for another example of using a linear equation in one variable to solve an application.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Solve a Coin Problem Using an Equation in One Variable\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/vOA7SGQCpr8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p>A cash register contains only five dollar and ten dollar bills. It contains twice as many five&#8217;s as ten&#8217;s and the total amount of money in the cash register is 560 dollars. How many ten&#8217;s are in the cash register?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q315017\">Show Solution<\/span><\/p>\n<div id=\"q315017\" class=\"hidden-answer\" style=\"display: none\">There are 28 ten dollar bills in the cash register.<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm13829\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=13829&theme=oea&iframe_resize_id=ohm13829&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>Many applications are solved using known formulas. The problem is stated, a formula is identified, the known quantities are substituted into the formula, the equation is solved for the unknown, and the problem\u2019s question is answered. Sometimes, these problems involve two equations representing two unknowns, which can be written using one equation in one variable by expressing one unknown in terms of the other. Examples of formulas include the <strong>perimeter<\/strong> of a rectangle, [latex]P=2L+2W[\/latex]; the <strong>area<\/strong> of a rectangular region, [latex]A=LW[\/latex]; and the <strong>volume<\/strong> of a rectangular solid, [latex]V=LWH[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Perimeter Problem<\/h3>\n<p>The perimeter of a rectangular outdoor patio is [latex]54[\/latex] ft. The length is [latex]3[\/latex] ft. greater than the width. What are the dimensions of the patio?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q815614\">Show Solution<\/span><\/p>\n<div id=\"q815614\" class=\"hidden-answer\" style=\"display: none\">\n<p>The perimeter formula is standard: [latex]P=2L+2W[\/latex]. We have two unknown quantities, length and width. However, we can write the length in terms of the width as [latex]L=W+3[\/latex]. Substitute the perimeter value and the expression for length into the formula. It is often helpful to make a sketch and label the sides as shown below.<\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200341\/CNX_CAT_Figure_02_03_003.jpg\" alt=\"A rectangle with the length labeled as: L = W + 3 and the width labeled as: W.\" width=\"487\" height=\"166\" \/><\/p>\n<p>Now we can solve for the width and then calculate the length.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}P=2L+2W\\hfill \\\\ 54=2\\left(W+3\\right)+2W\\hfill \\\\ 54=2W+6+2W\\hfill \\\\ 54=4W+6\\hfill \\\\ 48=4W\\hfill \\\\ 12=W\\hfill \\\\ \\left(12+3\\right)=L\\hfill \\\\ 15=L\\hfill \\end{array}[\/latex]<\/div>\n<p>The dimensions are [latex]L=15[\/latex] ft and [latex]W=12[\/latex] ft.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Evaluating a variable for an expression<\/h3>\n<p>In the example above, the length was written in terms of the width in order to substitute it for the variable [latex]L[\/latex] to be able to write one equation in one variable.<\/p>\n<p>This is a powerful technique in algebra and should be practiced until it becomes familiar, so it&#8217;s worth taking a closer look.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q866900\">more<\/span><\/p>\n<div id=\"q866900\" class=\"hidden-answer\" style=\"display: none\">\n<p>We had [latex]P = 2L + 2W[\/latex], which is one equation in two variables. Given the value for [latex]P[\/latex], we don&#8217;t have enough information to solve for either [latex]L[\/latex] or [latex]W[\/latex].<\/p>\n<p>But we do know that [latex]L=W+3[\/latex]. We can use this to substitute for the [latex]L[\/latex] in the original equation. Be careful to retain the original statement exactly, substituting only the value for [latex]L[\/latex]. A handy way to handle the substitution is with a set of parentheses.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}P=2L+2W\\hfill \\\\ 54=2\\left(L\\right)+2W \\,\\,\\,\\,\\,\\,\\,\\, \\text{ wrap the } L \\text{ to see where to make the substitution} \\hfill \\\\ 54=2\\left(W+3\\right)+2W \\,\\,\\,\\,\\, \\text{ drop the expression for } L \\text{ into the parentheses}\\hfill \\\\ 54=2W+6+2W\\hfill \\\\ 54=4W+6\\end{array}[\/latex]<\/p>\n<p>Now we have one equation in one variable. We can use the properties of equality to isolate the variable [latex]W[\/latex] on one side, with its numerical value on the other.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the dimensions of a rectangle given that the perimeter is [latex]110[\/latex] cm. and the length is 1 cm. more than twice the width.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q447289\">Show Solution<\/span><\/p>\n<div id=\"q447289\" class=\"hidden-answer\" style=\"display: none\">\n<p>L = 37 cm, W = 18 cm<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm7679\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7679&theme=oea&iframe_resize_id=ohm7679&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Area Problem<\/h3>\n<p>The perimeter of a tablet of graph paper is 48 in<sup>2<\/sup>. The length is [latex]6[\/latex] in. more than the width. Find the area of the graph paper.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q81698\">Show Solution<\/span><\/p>\n<div id=\"q81698\" class=\"hidden-answer\" style=\"display: none\">\n<p>The standard formula for area is [latex]A=LW[\/latex]; however, we will solve the problem using the perimeter formula. The reason we use the perimeter formula is because we know enough information about the perimeter that the formula will allow us to solve for one of the unknowns. As both perimeter and area use length and width as dimensions, they are often used together to solve a problem such as this one.<\/p>\n<p>We know that the length is 6 in. more than the width, so we can write length as [latex]L=W+6[\/latex]. Substitute the value of the perimeter and the expression for length into the perimeter formula and find the length.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}P=2L+2W\\hfill \\\\ 48=2\\left(W+6\\right)+2W\\hfill \\\\ 48=2W+12+2W\\hfill \\\\ 48=4W+12\\hfill \\\\ 36=4W\\hfill \\\\ 9=W\\hfill \\\\ \\left(9+6\\right)=L\\hfill \\\\ 15=L\\hfill \\end{array}[\/latex]<\/div>\n<p>Now, we find the area given the dimensions of [latex]L=15[\/latex] in. and [latex]W=9[\/latex] in.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}A\\hfill&=LW\\hfill \\\\ A\\hfill&=15\\left(9\\right)\\hfill \\\\ \\hfill&=135\\text{ in}^{2}\\hfill \\end{array}[\/latex]<\/div>\n<p>The area is [latex]135[\/latex] in<sup>2<\/sup>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>A game room has a perimeter of 70 ft. The length is five more than twice the width. How many ft<sup>2<\/sup> of new carpeting should be ordered?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q646790\">Show Solution<\/span><\/p>\n<div id=\"q646790\" class=\"hidden-answer\" style=\"display: none\">\n<p>250 ft<sup>2<\/sup><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm1688\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1688&theme=oea&iframe_resize_id=ohm1688&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\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-3897\">\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>Modification and Revision . <strong>Provided by<\/strong>: Lumen Learning. <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>College Algebra Corequisite. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\">https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Precalculus. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/precalculus\/\">https:\/\/courses.lumenlearning.com\/precalculus\/<\/a>. <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>\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":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Modification and Revision \",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra Corequisite\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/precalculus\/\",\"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-3897","chapter","type-chapter","status-publish","hentry"],"part":3090,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3897","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":7,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3897\/revisions"}],"predecessor-version":[{"id":4924,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3897\/revisions\/4924"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/3090"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3897\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=3897"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=3897"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=3897"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=3897"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}