최댓값·최솟값과 최적화

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.