최댓값·최솟값과 최적화

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.