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

We want to find the maximum possible area subject to the constraint that the total fencing is [latex]100\text{ft}.[/latex] From (Figure), the total amount of fencing used will be [latex]2x+y.[/latex] Therefore, the constraint equation is

Solving this equation for [latex]y,[/latex] we have [latex]y=100-2x.[/latex] Thus, we can write the area as

Before trying to maximize the area function [latex]A(x)=100x-2{x}^{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’s consider the function [latex]A(x)=100x-2{x}^{2}[/latex] over the closed interval [latex]\left[0,50\right].[/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:

Maximize [latex]A(x)=100x-2{x}^{2}[/latex] over the interval [latex]\left[0,50\right].[/latex]

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

Therefore, the only critical point is [latex]x=25[/latex] ((Figure)). 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\text{ft}.[/latex] The area of this garden is [latex]1250{\text{ft}}^{2}.[/latex]

Figure 2. To maximize the area of the garden, we need to find the maximum value of the function [latex]A(x)=100x-2{x}^{2}.[/latex]

Step 1: Let [latex]x[/latex] be the side length of the square to be removed from each corner ((Figure)). 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.

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.

Step 2: We are trying to maximize the volume of a box. Therefore, the problem is to maximize [latex]V.[/latex]

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·W·H,[/latex] where [latex]L,W,\text{ and }H[/latex] are the length, width, and height, respectively.

Step 4: From (Figure), 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

Step 5: To determine the domain of consideration, let’s examine (Figure). 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]\left[0,12\right],[/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]\left[0,12\right][/latex] and check whether the absolute maximum occurs at an interior point.

Step 6: Since [latex]V(x)[/latex] is a continuous function over the closed, bounded interval [latex]\left[0,12\right],[/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

To find the critical points, we need to solve the equation

Dividing both sides of this equation by 12, the problem simplifies to solving the equation

Using the quadratic formula, we find that the critical points are

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}\text{in}.[/latex] 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.

Figure 4. Maximizing the volume of the box leads to finding the maximum value of a cubic polynomial.

Step 1: Let [latex]x[/latex] be the distance running and let [latex]y[/latex] be the distance swimming ((Figure)). Let [latex]T[/latex] be the time it takes to get from the cabin to the island.

Figure 5. How can we choose [latex]x[/latex] and [latex]y[/latex] to minimize the travel time from the cabin to the island?

Step 2: The problem is to minimize [latex]T.[/latex]

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]×[/latex] Time [latex](D=R×T),[/latex] the time spent running is

and the time spent swimming is

Therefore, the total time spent traveling is

Step 4: From (Figure), the line segment of [latex]y[/latex] miles forms the hypotenuse of a right triangle with legs of length [latex]2\text{mi}[/latex] and [latex]6-x\text{mi}.[/latex] 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

Step 5: From (Figure), we see that [latex]0\le x\le 6.[/latex] Therefore, [latex]\left[0,6\right][/latex] is the domain of consideration.

Step 6: Since [latex]T(x)[/latex] is a continuous function over a closed, bounded interval, it has a maximum and a minimum. Let’s begin by looking for any critical points of [latex]T[/latex] over the interval [latex]\left[0,6\right].[/latex] The derivative is

If [latex]{T}^{\prime }(x)=0,[/latex] then

Therefore,

Squaring both sides of this equation, we see that if [latex]x[/latex] satisfies this equation, then [latex]x[/latex] must satisfy

which implies

We conclude that if [latex]x[/latex] is a critical point, then [latex]x[/latex] satisfies

Therefore, the possibilities for critical points are

Since [latex]x=6+6\text{/}\sqrt{55}[/latex] is not in the domain, it is not a possibility for a critical point. On the other hand, [latex]x=6-6\text{/}\sqrt{55}[/latex] is in the domain. Since we squared both sides of (Figure) to arrive at the possible critical points, it remains to verify that [latex]x=6-6\text{/}\sqrt{55}[/latex] satisfies (Figure). Since [latex]x=6-6\text{/}\sqrt{55}[/latex] does satisfy that equation, we conclude that [latex]x=6-6\text{/}\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=6-6\text{/}\sqrt{55}.[/latex] We find that [latex]T(0)\approx 2.108\text{h}[/latex] and [latex]T(6)\approx 1.417\text{h,}[/latex] whereas [latex]T(6-6\text{/}\sqrt{55})\approx 1.368\text{h}.[/latex] Therefore, we conclude that [latex]T[/latex] has a local minimum at [latex]x\approx 5.19[/latex] mi.

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.

Step 2: The problem is to maximize [latex]R.[/latex]

Step 3: The revenue (per day) is equal to the number of cars rented per day times the price charged per car per day—that is, [latex]R=n×p.[/latex]

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

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]\left[50,200\right].[/latex]

Step 6: Since [latex]R[/latex] is a continuous function over the closed, bounded interval [latex]\left[50,200\right],[/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.

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.

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.

Figure 7. We want to maximize the area of a rectangle inscribed in an ellipse.

Step 2: The problem is to maximize [latex]A.[/latex]

Step 3: The area of the rectangle is [latex]A=LW.[/latex]

Step 4: Let [latex](x,y)[/latex] be the corner of the rectangle that lies in the first quadrant, as shown in (Figure). 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{\frac{1-{x}^{2}}{4}}.[/latex] Therefore, the area is

Step 5: From (Figure), 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

Step 6: As mentioned earlier, [latex]A(x)[/latex] is a continuous function over the closed, bounded interval [latex]\left[0,2\right].[/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]00.[/latex] Therefore, the maximum must occur at a critical point. Taking the derivative of [latex]A(x),[/latex] we obtain

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

then [latex]x[/latex] must satisfy

Therefore, [latex]{x}^{2}=2.[/latex] Thus, [latex]x=\text{±}\sqrt{2}[/latex] are the possible solutions of (Figure). Since we are considering [latex]x[/latex] over the interval [latex]\left[0,2\right],[/latex] [latex]x=\sqrt{2}[/latex] is a possibility for a critical point, but [latex]x=\text{−}\sqrt{2}[/latex] is not. Therefore, we check whether [latex]\sqrt{2}[/latex] is a solution of (Figure). Since [latex]x=\sqrt{2}[/latex] is a solution of (Figure), we conclude that [latex]\sqrt{2}[/latex] is the only critical point of [latex]A(x)[/latex] in the interval [latex]\left[0,2\right].[/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

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=LW=(2\sqrt{2})(\sqrt{2})=4.[/latex]

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.

Figure 8. We want to minimize the surface area of a square-based box with a given volume.

Step 2: We need to minimize the surface area. Therefore, we need to minimize [latex]S.[/latex]

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·y.[/latex] The area of the base is [latex]{x}^{2}.[/latex] Therefore, the surface area of the box is

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

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:

Therefore, [latex]S(x)=\frac{864}{x}+{x}^{2}.[/latex]

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]

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

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)). When [latex]x=6\sqrt[3]{2},[/latex] [latex]y=\frac{216}{{(6\sqrt[3]{2})}^{2}}=3\sqrt[3]{2}\text{in}.[/latex] Therefore, the dimensions of the box should be [latex]x=6\sqrt[3]{2}\text{in}.[/latex] and [latex]y=3\sqrt[3]{2}\text{in}.[/latex] With these dimensions, the surface area is

Figure 9. We can use a graph to determine the dimensions of a box of given the volume and the minimum surface area.