We need to maximize the function [latex]A(x)=200x-2x^2[/latex] over the interval [latex][0,100][/latex].

The maximum area is [latex]5000 \, \text{ft}^2[/latex].

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.

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 \cdot W \cdot H[/latex], where [latex]L, \, W[/latex], and [latex]H[/latex] are the length, width, and height, respectively.

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,

Step 5: To determine the domain of consideration, let’s 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.

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

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

Dividing both sides of this equation by [latex]12[/latex], 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}[/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.

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.

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

and the time spent swimming is,

Therefore, the total time spent traveling is,

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

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.

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][0,6][/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+\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.

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 \times 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][50,200][/latex].

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.

Watch the following video to see the worked solution to this example.

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.

You can view the transcript for this segmented clip of “4.7 Applied Optimization Problems” here (opens in new window).

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.

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

Step 3: The area of the rectangle is [latex]A=L \times W[/latex].

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:

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

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.

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]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=\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=−\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,

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].