# What is the maximum volume of the box, given the parameters below?

## Given a rectangular sheet of cardboard, $16$ in. by $10$ in., you are asked to cut off identical squares from each of the four corners of the sheet and then bend up the sides of the remaining cardboard to form a rectangular box.

Dec 20, 2016

$x = 2$ results in the maximum volume.

The associated max volume is given by:

$V = 144$ cubic inches

#### Explanation:

Let us set up the following variables:

$\left\{\begin{matrix}w & \text{Width of the Box (in)" \\ l & "Length of the Box (in)" \\ x & "Length of the Corner Cut-out (in)" \\ V & "Volume of the Box (cubic in)}\end{matrix}\right.$

We want to vary the corner length $x$ such that we maximise $V$, ie find a critical point of $\frac{\mathrm{dV}}{\mathrm{dx}}$ that is a maximum, so we to find a function $V = V \left(x\right)$.

The dimensions of the sheet are 16" by 10", hence, wlog (without loss of generality) taking $w = 16$ we get:

Width: $x + w + x = 16 \setminus \setminus \implies w = 16 - 2 x = 2 \left(8 - x\right)$
Length: $x + l + x = 10 \implies l = 10 - 2 x = 2 \left(5 - x\right)$

Then the volume is given by:

$\setminus \setminus \setminus \setminus \setminus V = w l x$
$\therefore V = 2 \left(8 - x\right) 2 \left(5 - x\right) x$
$\therefore V = 4 x \left({x}^{2} - 13 x + 40\right)$
$\therefore V = 4 {x}^{3} - 52 {x}^{2} + 160 x$

Differentiating wrt $x$ gives us;

$\therefore \frac{\mathrm{dV}}{\mathrm{dx}} = 12 {x}^{2} - 104 x + 160$

At a critical point, $\frac{\mathrm{dV}}{\mathrm{dx}} = 0$

$\therefore 12 {x}^{2} - 104 x + 160 = 0$
$\therefore 3 {x}^{2} - 26 x + 40 = 0$
$\therefore \left(3 x - 20\right) \left(x - 2\right) = 0$
$\therefore x = 2 , \frac{20}{3} = 0$

We should check while value leads to a maximum volume

$\therefore \frac{{d}^{2} V}{\mathrm{dx}} ^ 2 = 24 x - 104$

$x = 2 \implies \frac{{d}^{2} V}{\mathrm{dx}} ^ 2 < 0 \implies \max$
$x = \frac{20}{3} \implies \frac{{d}^{2} V}{\mathrm{dx}} ^ 2 > 0 \implies \min$

So we have a maximum volume when $x = 2$, the corresponding volume is:

$V = 4 \cdot 8 - 52 \cdot 4 + 160 \cdot 2 = 32 - 208 + 320 = 144$

If we graph the Volume Function, $V \left(x\right) = 4 {x}^{3} - 52 {x}^{2} + 160 x$ we can confirm our result:
graph{4x^3-52x^2+160x [-3, 10, -70, 160]}
Hopefully you can visually confirm the above #