Question

A closed box, whose length in twice its width, is to have a surface area of 192 square inches. Find the dimensions of the box when the volume is maximum?

Answer 1

Let $l,w,h$ be the dimensions of the closed box.

Given: $l=2w$ and Surface area $=192$ square inches.

$\Rightarrow 2lw+2wh+2lh=192$

$\Rightarrow lw+wh+lh=96$

$\Rightarrow2w^2+wh+2wh=96$

$\Rightarrow2w^2+3wh=96$

$\Rightarrow h=\frac{96-2w^2}{3w}$ ... (1)

Now volume of the box is $V=lwh$

$\Rightarrow V=(lw)\frac{96-2w^2}{3w}$ ( from (1) )

$\Rightarrow V=\frac{96l-2lw^2}{3}$

Let $f(l,w)=\frac{96l-2lw^2}{3}$

Find the first order partial derivatives of the above function.

$f(l,w)=\frac{96l-2lw^2}{3}\Rightarrow f_{l}=\frac{96-2w^2}{3}$ and $f_{w}=\frac{0-4lw}{3}$

$\Rightarrow f_{l}=\frac{96-2w^2}{3}$ and $f_{w}=\frac{-4lw}{3}$

For maximum volume, set $f_{l}=0$ and $f_{w}=0$

$\Rightarrow \frac{96-2w^2}{3}=0$ and $\frac{-4lw}{3}=0$

$\Rightarrow 2w^2=96$ and $l=0$ or $w=0$

$\Rightarrow w^2=48$ ( since $l\neq 0,w\neq 0$ )

$\Rightarrow w=4\sqrt{3}$

Substituting $w=4\sqrt{3}$ in equation (1), we get $h=0$

And $w=4\sqrt{3}\Rightarrow l=8\sqrt{3}$

Therefore, dimensions of the closed box are: $l=8\sqrt{3}$ inches and $w=4{\sqrt{3}}$ inches