4.6 Applied Optimization Problems

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 4.39). Given $100$ ft of wire fencing, determine the dimensions that would create a garden of maximum area. What is the maximum area?

Maximize $A(x)=100x-2x^2$ over the interval $[0,50]$.

As mentioned earlier, since $A$ 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, $A(x)=0$. Since the area is positive for all $x$ in the open interval $(0,50)$, the maximum must occur at a critical point. Differentiating the function $A(x)$, we obtain

$$A^{\prime }(x)=100-4x.$$

Therefore, the only critical point is $x=25$ (Figure 4.40). We conclude that the maximum area must occur when $x=25$. Then we have $y=100-2x=100-2(25)=50$. To maximize the area of the garden, let $x=25$ ft and $y=50\mathrm{f}\mathrm{t}$. The area of this garden is $1250\mathrm{f}{\mathrm{t}}^2$.

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?

Step 2: We are trying to maximize the volume of a box. Therefore, the problem is to maximize $V$.

Step 3: As mentioned in step $2$, are trying to maximize the volume of a box. The volume of a box is $V=L\cdot W\cdot H$, where $L$, $W$, and $H$ are the length, width, and height, respectively.

Step 4: From Figure 4.41, we see that the height of the box is $x$ inches, the length is $36-2x$ inches, and the width is $24-2x$ inches. Therefore, the volume of the box is

$$V(x)=(36-2x)(24-2x)x=4x^3-120x^2+864x.$$

Step 5: To determine the domain of consideration, let’s examine Figure 4.41. Certainly, we need $x>0$. 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 $x$ over the open interval $(0,12)$. Since $V$ is a continuous function over the closed interval $[0,12]$, we know $V$ will have an absolute maximum over the closed interval. Therefore, we consider $V$ over the closed interval $[0,12]$ and check whether the absolute maximum occurs at an interior point.

Step 6: Since $V(x)$ is a continuous function over the closed, bounded interval $[0,12]$, $V$ must have an absolute maximum (and an absolute minimum). Since $V(x)=0$ at the endpoints and $V(x)>0$ for , the maximum must occur at a critical point. The derivative is

$$V^{\prime }(x)=12x^2-240x+864.$$

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

$$12x^2-240x+864=0.$$

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

$$x^2-20x+72=0.$$

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

$$x=\frac{20\pm \sqrt{{(-20)}^2-4(1)(72)}}{2}=\frac{20\pm \sqrt{112}}{2}=\frac{20\pm 4\sqrt{7}}{2}=10\pm 2\sqrt{7}.$$

Since $10+2\sqrt{7}$ is not in the domain of consideration, the only critical point we need to consider is $10-2\sqrt{7}$. Therefore, the volume is maximized if we let $x=10-2\sqrt{7}\mathrm{in}$. The maximum volume is $V(10-2\sqrt{7})=640+448\sqrt{7}\approx 1825\mathrm{i}\mathrm{n}{.}^3$ as shown in the following graph.

An island is $2\mathrm{mi}$ due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that is $6\mathrm{mi}$ west of that point. The visitor is planning to go from the cabin to the island. Suppose the visitor runs at a rate of $8\mathrm{mph}$ and swims at a rate of $3\mathrm{mph}$. How far should the visitor run before swimming to minimize the time it takes to reach the island?

Step 2: The problem is to minimize $T$.

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 $=$ Rate $\times$ Time $(D=R\times T)$, the time spent running is

$$T_{\text{running}}=\frac{D_{\text{running}}}{R_{\text{running}}}=\frac{x}{8},$$

and the time spent swimming is

$$T_{\text{swimming}}=\frac{D_{\text{swimming}}}{R_{\text{swimming}}}=\frac{y}{3}.$$

Therefore, the total time spent traveling is

$$T=\frac{x}{8}+\frac{y}{3}.$$

Step 4: From Figure 4.43, the line segment of $y$ miles forms the hypotenuse of a right triangle with legs of length $2\mathrm{mi}$ and $6-x\mathrm{mi}$. Therefore, by the Pythagorean theorem, $2^2+{(6-x)}^2=y^2$, and we obtain $y=\sqrt{{(6-x)}^2+4}$. Thus, the total time spent traveling is given by the function

$$T(x)=\frac{x}{8}+\frac{\sqrt{{(6-x)}^2+4}}{3}.$$

Step 5: From Figure 4.43, we see that $0\le x\le 6$. Therefore, $[0,6]$ is the domain of consideration.

Step 6: Since $T(x)$ 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 $T$ over the interval $[0,6]$. The derivative is

$$T^{\prime }(x)=\frac{1}{8}-\frac{1}{2}\frac{{[{(6-x)}^2+4]}^{-1/2}}{3}\cdot 2(6-x)=\frac{1}{8}-\frac{(6-x)}{3\sqrt{{(6-x)}^2+4}}.$$

If $T^{\prime }(x)=0$, then

$$\frac{1}{8}=\frac{6-x}{3\sqrt{{(6-x)}^2+4}}.$$

Therefore,

$$3\sqrt{{(6-x)}^2+4}=8(6-x).$$

(4.3)

Squaring both sides of this equation, we see that if $x$ satisfies this equation, then $x$ must satisfy

$$9[{(6-x)}^2+4]=64{(6-x)}^2,$$

which implies

$$55{(6-x)}^2=36.$$

We conclude that if $x$ is a critical point, then $x$ satisfies

$${(x-6)}^2=\frac{36}{55}.$$

Therefore, the possibilities for critical points are

$$x=6\pm \frac{6}{\sqrt{55}}.$$

Since $x=6+6/\sqrt{55}$ is not in the domain, it is not a possibility for a critical point. On the other hand, $x=6-6/\sqrt{55}$ is in the domain. Since we squared both sides of Equation 4.3 to arrive at the possible critical points, it remains to verify that $x=6-6/\sqrt{55}$ satisfies Equation 4.3. Since $x=6-6/\sqrt{55}$ does satisfy that equation, we conclude that $x=6-6/\sqrt{55}$ is a critical point, and it is the only one. To justify that the time is minimized for this value of $x$, we just need to check the values of $T(x)$ at the endpoints $x=0$ and $x=6$, and compare them with the value of $T(x)$ at the critical point $x=6-6/\sqrt{55}$. We find that $T(0)\approx 2.108\mathrm{hr}$ and $T(6)\approx 1.417\mathrm{hr},$ whereas $T\left(6-6/\sqrt{55}\right)\approx 1.368\mathrm{hr}$. Therefore, we conclude that $T$ has a local minimum at $x\approx 5.19\mathrm{mi}$.

Owners of a car rental company have determined that if they charge customers $p$ dollars per day to rent a car, where $50\le p\le 200$, the number of cars $n$ they rent per day can be modeled by the linear function $n(p)=1000-5p$. If they charge $\$50$ per day or less, they will rent all their cars. If they charge $\$200$ per day or more, they will not rent any cars. Assuming the owners plan to charge customers between $50 per day and $\$200$ per day to rent a car, how much should they charge to maximize their revenue?

Step 2: The problem is to maximize $R$.

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, $R=n\times p$.

Step 4: Since the number of cars rented per day is modeled by the linear function $n(p)=1000-5p$, the revenue $R$ can be represented by the function

$$R(p)=n\times p=(1000-5p)p=-5p^2+1000p.$$

Step 5: Since the owners plan to charge between $\$50$ per car per day and $\$200$ per car per day, the problem is to find the maximum revenue $R(p)$ for $p$ in the closed interval $[50,200]$.

Step 6: Since $R$ is a continuous function over the closed, bounded interval $[50,200]$, it has an absolute maximum (and an absolute minimum) in that interval. To find the maximum value, look for critical points. The derivative is $R^{\prime }(p)=-10p+1000$. Therefore, the critical point is $p=100$ When $p=100$, $R(100)=\$50,000$. When $p=50$, $R(p)=\$37,500$. When $p=200$, $R(p)=\$0$. Therefore, the absolute maximum occurs at $p=\$100$. The car rental company should charge $\$100$ per day per car to maximize revenue as shown in the following figure.

A rectangle is to be inscribed in the ellipse

$$\frac{x^2}{4}+y^2=1.$$

What should the dimensions of the rectangle be to maximize its area? What is the maximum area?

Step 2: The problem is to maximize $A$.

Step 3: The area of the rectangle is $A=LW$.

Step 4: Let $(x,y)$ be the corner of the rectangle that lies in the first quadrant, as shown in Figure 4.45. We can write length $L=2x$ and width $W=2y$. Since ${\displaystyle \frac{x^2}{4}}+y^2=1$ and $y>0$, we have $y=\sqrt{1-{\displaystyle \frac{x^2}{4}}}$. Therefore, the area is

$$A=LW=\left(2x\right)\left(2y\right)=4x\sqrt{1-\frac{x^2}{4}}=2x\sqrt{4-x^2}.$$

Step 5: From Figure 4.45, we see that to inscribe a rectangle in the ellipse, the $x$-coordinate of the corner in the first quadrant must satisfy . Therefore, the problem reduces to looking for the maximum value of $A(x)$ over the open interval $(0,2)$. Since $A(x)$ will have an absolute maximum (and absolute minimum) over the closed interval $[0,2]$, we consider $A(x)=2x\sqrt{4-x^2}$ over the interval $[0,2]$. 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, $A(x)$ is a continuous function over the closed, bounded interval $[0,2]$. Therefore, it has an absolute maximum (and absolute minimum). At the endpoints $x=0$ and $x=2$, $A(x)=0$. For , $A(x)>0$. Therefore, the maximum must occur at a critical point. Taking the derivative of $A(x)$, we obtain

	$A^{\prime }(x)$	$=2\sqrt{4-x^2}+2x\cdot {\displaystyle \frac{1}{2\sqrt{4-x^2}}}(-2x)$
		$=2\sqrt{4-x^2}-{\displaystyle \frac{2x^2}{\sqrt{4-x^2}}}$
		$={\displaystyle \frac{8-4x^2}{\sqrt{4-x^2}}}.$

To find critical points, we need to find where $A^{\prime }(x)=0$. We can see that if $x$ is a solution of

$$\frac{8-4x^2}{\sqrt{4-x^2}}=0,$$

(4.4)

then $x$ must satisfy

$$8-4x^2=0.$$

Therefore, $x^2=2$. Thus, $x=\pm \sqrt{2}$ are the possible solutions of Equation 4.4. Since we are considering $x$ over the interval $[0,2]$, $x=\sqrt{2}$ is a possibility for a critical point, but $x=-\sqrt{2}$ is not. Therefore, we check whether $\sqrt{2}$ is a solution of Equation 4.4. Since $x=\sqrt{2}$ is a solution of Equation 4.4, we conclude that $\sqrt{2}$ is the only critical point of $A(x)$ in the interval $[0,2]$. Therefore, $A(x)$ must have an absolute maximum at the critical point $x=\sqrt{2}$. To determine the dimensions of the rectangle, we need to find the length $L$ and the width $W$. If $x=\sqrt{2}$ then

$$y=\sqrt{1-\frac{{\left(\sqrt{2}\right)}^2}{4}}=\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}.$$

Therefore, the dimensions of the rectangle are $L=2x=2\sqrt{2}$ and $W=2y={\displaystyle \frac{2}{\sqrt{2}}}=\sqrt{2}$. The area of this rectangle is $A=LW=\left(2\sqrt{2}\right)\left(\sqrt{2}\right)=4$.

A rectangular box with a square base, an open top, and a volume of $216$ in.³ is 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?

Step 2: We need to minimize the surface area. Therefore, we need to minimize $S$.

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 $x\cdot y$. The area of the base is $x^2$. Therefore, the surface area of the box is

$$S=4xy+x^2.$$

Step 4: Since the volume of this box is $x^{2}y$ and the volume is given as $216\mathrm{in}{.}^3$, the constraint equation is

$$x^{2}y=216.$$

Solving the constraint equation for $y$, we have $y={\displaystyle \frac{216}{x^2}}$. Therefore, we can write the surface area as a function of $x$ only:

$$S(x)=4x\left(\frac{216}{x^2}\right)+x^2.$$

Therefore, $S(x)={\displaystyle \frac{864}{x}}+x^2$.

Step 5: Since we are requiring that $x^{2}y=216$, we cannot have $x=0$. Therefore, we need $x>0$. On the other hand, $x$ is allowed to have any positive value. Note that as $x$ becomes large, the height of the box $y$ becomes correspondingly small so that $x^{2}y=216$. Similarly, as $x$ becomes small, the height of the box becomes correspondingly large. We conclude that the domain is the open, unbounded interval $(0,\mathrm{\infty })$. 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 $(0,\mathrm{\infty })$.

Step 6: Note that as $x\to 0^{+}$, $S(x)\to \mathrm{\infty }$. Also, as $x\to \mathrm{\infty }$, $S(x)\to \mathrm{\infty }$. Since $S$ is a continuous function that approaches infinity at the ends, it must have an absolute minimum at some $x\in (0,\mathrm{\infty })$. This minimum must occur at a critical point of $S$. The derivative is

$$S^{\prime }(x)=-\frac{864}{x^2}+2x.$$

Therefore, $S^{\prime }(x)=0$ when $2x={\displaystyle \frac{864}{x^2}}$. Solving this equation for $x$, we obtain $x^3=432$, so $x=\sqrt[3]{432}=6\sqrt[3]{2}$. Since this is the only critical point of $S$, the absolute minimum must occur at $x=6\sqrt[3]{2}$ (see Figure 4.47). When $x=6\sqrt[3]{2}$, $y={\displaystyle \frac{216}{{\left(6\sqrt[3]{2}\right)}^2}}=3\sqrt[3]{2}\mathrm{in}$. Therefore, the dimensions of the box should be $x=6\sqrt[3]{2}\mathrm{in}$. and $y=3\sqrt[3]{2}\mathrm{in}$. With these dimensions, the surface area is

$$S\left(6\sqrt[3]{2}\right)=\frac{864}{6\sqrt[3]{2}}+{\left(6\sqrt[3]{2}\right)}^2=108\sqrt[3]{4}\mathrm{in}{.}^2$$