# What will the dimensions of the resulting cardboard box be if the company wants to maximize the volume and they start with a flat piece of square cardboard 20 feet per side, and then cut smaller squares out of each corner and fold up the sides to create the box?

Mar 24, 2015

Suppose that the squares removed from each corner are $x$ feet by $x$ feet each.

When these are folded up they give a box with a height of $x$ feet
and a base of $20 - 2 x$ feet by $20 - 2 x$ feet
for a volume
$V = x {\left(20 - 2 x\right)}^{2} = 400 x - 80 {x}^{2} + 4 {x}^{3}$

To find the critical point(s) take the derivative of $V$, set it to zero, and solve for $x$.

$\frac{\mathrm{dV}}{\mathrm{dx}} = 400 - 160 x + 12 {x}^{2}$

$= 4 \left(3 x - 10\right) \left(x - 10\right) = 0$

Since $x = 10$ gives a Volume of $0$

$\rightarrow$ the critical point for the Volume that is it's maximum occurs when $x = \frac{10}{3}$.

The resulting box will be
$3 \frac{1}{3} \times 13 \frac{1}{3} \times 13 \frac{1}{3}$ feet