최댓값·최솟값과 최적화

$h=\frac{62}{3}$ in.

Call the length of the wire $L$ units and suppose that it is cut $\ell$ units from one end. Make the square from the piece of length $\ell$, and make the circle from the remaining piece of length $L-\ell$.

The square has perimeter $\ell$, so its side length is $\ell/4$ and its area is $\left(\dfrac{\ell}{4}\right)^2$. The circle has circumference $L-\ell$, so its radius is $\dfrac{L-\ell}{2\pi}$ and its area is $\pi\left(\dfrac{L-\ell}{2\pi}\right)^2=\dfrac{(L-\ell)^2}{4\pi}$.

The area enclosed by the shapes, when the square is made from a length of size $\ell$, is

$$\begin{aligned} A(\ell)&=\dfrac{\ell^2}{16}+\dfrac{(L-\ell)^2}{4\pi} \\ \text{We want to find the global max and min for this function, given the constraint $0 \leq \ell \leq L$, so we find its derivative:} \\ A'(\ell)&=\dfrac{\ell}{8} -\dfrac{L-\ell}{2\pi} =\dfrac{\pi+4}{8\pi}\ell-\dfrac{L}{2\pi} \\ \text{Now, we find the critical point.} \\ A'(\ell)&=0\\ \frac{\pi+4}{8\pi}\ell&=\frac{L}{2\pi}\\ \ell&=\frac{4L}{\pi+4} \end{aligned}$$

$$\begin{array}{|c||c|c|c|}\hline \ell & 0 & L & \dfrac{.^.}{.}\frac{4L}{\pi+4} \\ \hline \text{type} & \text{endpoint} & \text{endpoint} & \text{critical point} \\ \hline A(\ell) & \dfrac{.^.}{.}\frac{L^2}{4\pi} & \frac{L^2}{16}_{} & A\left(\frac{4L}{\pi+4}\right) \\ \hline \end{array}$$

It seems obnoxious to evaluate $A\left(\frac{4L}{\pi+4}\right)$, and the problem doesn't ask for it–but we still have to figure out whether it is a global max or min.

When $\ell<\frac{4L}{\pi+4}$, $A'(\ell)<0$, and when $\ell>\frac{4L}{\pi+4}$, $A'(\ell)>0$. So, $A(\ell)$ is decreasing until $\ell=\frac{4L}{\pi+4}$, then increasing. That means our critical point $\ell=\frac{4L}{\pi+4}$ is a local minimum.

So, the minimum occurs at the only critical point, which is $\ell=\left(\dfrac{4}{4+\pi}\right)L$. This corresponds to $\dfrac{\ell}{L}=\dfrac{4}{4+\pi}$: the proportion of the wire that is cut is $\dfrac{4}{4+\pi}$.

The maximum has to be either at $\ell=0$ or at $\ell=L$. As $A(0)=\dfrac{L^2}{4\pi} > A(L)=\dfrac{L^2}{16}$, the maximum has $\ell=0$ (that is, no square).

Let $x$ be the length of one side of the square bottom and $h$ the height of one of the sides. We first note that the volume, $V$, of the box is $V = x^2 h$, and the surface area of the box is $S = 2x^2 + 4xh$, since there are two square sides (bottom and top) of area $x^2$ and four identical sides of area $xh$. Since the material for the top and bottom costs $3.00 per square foot and the material for the sides costs $1.50 per square foot, it follows that the total cost of a box with surface area $S$ is

$$C = 3.00 (2x^2) + 1.50 (4xh) = 6x^2 + 6xh\,.$$

We are interested in building box of largest volume for which we spend $15. This cost constraint implies that

$$15 = 6x^2 + 6xh\,,$$ which allows us to express $h$ as a function of $x$ by solving for $h$. Doing so, we see that $6xh = 15 - 6x^2$, so

$$h = \frac{15-6x^2}{6x}\,.$$ Using this relationship, we are now able to express the volume, $V$, of the box as a function of the single variable $x$. Subsituting the expression for $h$, we have

$$V = x^2 h = x^2 \cdot \frac{15-6x^2}{6x} = \frac{1}{6}x(15 - 6x^2) = \frac{15}{6}x - x^3\,.$$ Note particularly that in order to have a box, we require that both $x > 0$ and $h > 0$. The latter condition along with the fact that $h = \frac{15-6x^2}{6x}$ shows that we need $15 - 6x^2 > 0$, and thus $x^2 < \frac{15}{6}$, so $x < \sqrt{\frac{15}{6}}$. Hence we are considering $V(x)$ on the domain $0 < x < \sqrt{\frac{15}{6}}$.

It is apparent that the box vanishes (and its volume tends to zero) as either $x \to 0^+$ or $x \to \sqrt{\frac{15}{6}}^-$, and thus the absolute maximum volume must appear between these values at a critical number. Differentiating $V$, we find that $V'(x) = \frac{15}{6} - 3x^2$. Solving $V'(x) = 0$ to find all critical numbers of $V$ in the domain, we see that $x^2 = \frac{15}{18}$, and thus $x = \pm \sqrt{\frac{15}{18}}$, so only $x = \sqrt{\frac{15}{18}}$ is a critical number that lies in $0 < x < \sqrt{\frac{15}{6}}$. Noting that the absolute minimum volume $V = 0$ occurs at each endpoint, it must be the case that the absolute maximum volume is $V\left( \sqrt{\frac{15}{18}} \right) = \frac{15}{6}\left( \sqrt{\frac{15}{18}} \right) - \left( \sqrt{\frac{15}{18}} \right)^3 \approx 1.52145$ cubic feet.

We start by constructing a careful diagram of the overall situation. Let $x$ be the distance from the first pole to the second pole where the cable touches the ground, and let $y$ be the length of the cable from the top of the first tower to the ground, and $z$ the length of cable from the ground to the top of the second tower, as pictured in the following figure. Note that we have included the given heights of the towers and the distance between them.

The total amount of cable, $L$, used (which we wish to minimize) is given by $L = y + z$. We can use the geometry of the situation to write both $y$ and $z$ in terms of $x$. In the first right triangle, by the Pythagorean Theorem we know that $x^2 + 60^2 = y^2$, and thus $y = \sqrt{x^2 + 3600}$. In the second right triangle, note that the horizontal leg of the triangle has length $100-x$, and thus $(100-x)^2 + 80^2 = z^2$, so $z = \sqrt{(100-x)^2 + 6400}$. It follows that with $L = y + z$,

$$L(x) = \sqrt{x^2 + 3600} + \sqrt{6400 + (100-x)^2}\,.$$ It is also apparent from the given situation that the relevant domain for $L$ is $0 \le x \le 100$. Given that we want to minimize $L$ on this closed interval, it remains to find any critical numbers of $L$.

We note by the chain rule that

$$L'(x) = \frac{2x}{2(x^2+3600)^{1/2}} + \frac{-2(100-x)}{2((100-x)^2 + 6400)^{1/2}}$$

We use technology to solve the equation $L'(x) = 0$ and find that it has a single solution in the interval $0 \le x \le 100$: $x \approx 42.85714$.

To find the absolute minimum of $L$ on the relevant domain, we evaluate $L$ at the endpoints and the only critical number. Doing so, we fine $L(0) \approx 188.062$, $L(100) \approx 196.619$, and $L(42.85714) \approx 172.047$. Hence the least amount of cable we can use is approximately $172.047$ feet and is accomplished by anchoring the cable about $42.85714$ feet from the first tower.