# A box has a bottom with one edge 7 times as long as the other. If the box has no top and the volume is fixed at V, what dimensions minimize the surface area?

Let $x$ be the length of the shorter edge of the bottom of the box so that $7 x$ is the length of the longer edge. Let $h$ be the height of the box. The volume of the box is $V = {x}^{2} h$ and the surface area, excluding the top, is $S = 7 {x}^{2} + 2 x h + 14 x h = 7 {x}^{2} + 16 x h$.
Since the volume $V$ is fixed, we can solve for $h$ in terms of $x$ to get $h = \frac{V}{{x}^{2}}$ and the substitute this into the equation for the surface area to get $S$ as a function of $x$: $S = f \left(x\right) = 7 {x}^{2} + \frac{16 V}{x}$ for $x > 0$.
We want to minimize $S = f \left(x\right)$ for $x > 0$, so take its derivative to get $\frac{\mathrm{dS}}{\mathrm{dx}} = f ' \left(x\right) = 14 x - \frac{16 V}{x} ^ 2$. Now solve for any positive critical points by setting this equal to zero to get the equation $14 x = \frac{16 V}{x} ^ 2$ or ${x}^{3} = \frac{8 V}{7}$ and $x = \frac{2 {V}^{\frac{1}{3}}}{7} ^ \left(\frac{1}{3}\right) \setminus \approx 1.04552 {V}^{\frac{1}{3}}$.
That this is the unique point that minimizes the surface area follows since $\frac{{d}^{2} S}{{\mathrm{dx}}^{2}} = f ' ' \left(x\right) = 14 + \frac{32 V}{{x}^{3}} > 0$ for all $x > 0$, making the graph of $S = f \left(x\right)$ concave up for all $x > 0$.
The dimensions that minimize the volume are therefore $x = \frac{2 {V}^{\frac{1}{3}}}{7} ^ \left(\frac{1}{3}\right) \setminus \approx 1.04552 {V}^{\frac{1}{3}}$, $7 x = 2 \cdot {7}^{\frac{2}{3}} {V}^{\frac{1}{3}} \setminus \approx 7.31861 {V}^{\frac{1}{3}}$, and $h = \frac{V}{{x}^{2}} = \frac{{7}^{\frac{2}{3}}}{4} {V}^{\frac{1}{3}} \setminus \approx 0.91483 {V}^{\frac{1}{3}}$ and the minimum surface area is $S = f \left(\frac{2 {V}^{\frac{1}{3}}}{7} ^ \left(\frac{1}{3}\right)\right) = 12 \cdot {7}^{\frac{1}{3}} {V}^{\frac{2}{3}} \setminus \approx 22.95517 {V}^{\frac{2}{3}}$.