Minimum area of the inside of a parallelepiped

Question

A container with parallelepiped shape, empty inside, and square base, has capacity of 32 litres. We want to cover the inside of the container (without the top surface), with a sheet. What are:

• the sides $x$
• the height $h$

of that parallelepiped if the area of sheet must be minimum?

MY ATTEMPT

My attempt was to use these two equations:

$x^2h=0.032$ (it's the area of the parallelepiped in meters squared)

$D(4xh+x^2)=0$ (the area of the sheet must be the equal to the 5 faces of the parallelepiped, and I must find its minimum, thus the derivative equal to zero)

I then obtain, from the first equation, $h=\frac{0.032}{x^2}$ and then I substitute this value of $h$ in $4xh+x^2$, obtaining $\frac{0.128}{x}+x^2$. Then I do the derivative of this last expression, obtaining $\frac{0.128}{x^2}+2x$ and put $=0$ to find the minimum, obtaining $x^3=\frac{0.128}{2}$

DRAWING OF THE PROBLEM

Note that the face of the parallelepiped with the yellow diagonal lines is the one we shouldn't consider, since it is the removed top surface. The faces with the red arrow are the one with area $xh$, the base (orange) has area $x^2$ since it's a square

Answer 1

Just cross-checked with GeoGebra, you're correct!

(The $\color{#0066BB}{blue}$ line is $y=\frac{0.128}x+x^2$, the $\color{#DD8800}{orange}$ line is its derivative)