Question

A box with a square base and an open top is to have volume 62.5 m^3. Neglect the thickness of the material used to make the box and find the dimension that will minimize the amount of material used.

Answer 1

Let $x$ be the length of the sides of the base and $h$ be the height. Then,

We need to maximize;

$\displaystyle SA=\text{area of base+4(area of vertical side)}=x^2+4xh$

subject to;

$V=\text{(area of base)(height)}=x^2h$

Now, $\displaystyle V=\text{(area of base)(height)}=x^2h\Rightarrow 62.5=x^2h\Rightarrow h=\frac{62.5}{x^2}$

$\displaystyle \Rightarrow SA=x^2+4xh=x^2+4x\left(\frac{62.5}{x^2}\right)=x^2+250x^{-1}$

Thus, taking the derivative of $\displaystyle SA=x^2+250x^{-1}$ wrt $x$ yields;

$(SA)\prime=2x-250x^{-2}$

Setting $(SA)\prime=0$ yields;

$2x-250x^{-2}=0\Rightarrow x=5\ m$

The second derivative test verifies that $SA$ has a minimum at $x=5\ m$ since:

$\displaystyle (SA)\prime\prime=2+500x^{-3}=6$ which is positive at $x=5\ m$

Thus, at $x=5,\ SA=5^2+\frac{250}{5}=75$

Hence, the minimum surface area;

$SA=75\ m^2$, and

the box should have base $5\ m$ by $5\ m$ and height $2.5\ m$.