# The corners are removed from a sheet of paper that is 3 ft square. The sides are folded up to form an open square box. What is the maximum volume of the box?

Jul 23, 2017

Maximum volume is $2 \setminus f {t}^{3}$

#### Explanation:

Let us set up the following variables:

$\left\{\begin{matrix}x & \text{Height of box (ft)" \\ y & "Width of box (ft)" \\ V & "Volume of the box ("ft^3")}\end{matrix}\right.$

The box has an open top and is square, so it consists of a base, and four identical sides.

Then the width (and therefore length) of the box is given by:

$x + y + x = 3 \implies y = 3 - 2 x$

Note that one really important constraint on $x$ and $y$ is that:

$x > 0$
$y > 0 \implies 3 - 2 x > 0 \implies x < \frac{3}{2}$

Thus we must have $0 < x < \frac{3}{2}$, Without this constraint the box cannot physically be constructed! The Volume of the box is then:

$V =$(area of base) $\times$ (height)
$\setminus \setminus = y \cdot y \cdot x$
$\setminus \setminus = x {y}^{2}$
$\setminus \setminus = x {\left(3 - 2 x\right)}^{2}$
$\setminus \setminus = 4 {x}^{3} - 12 {x}^{2} + 9 x$

If we graph this volume function we get $\left(\text{remember } 0 < x < \frac{3}{2}\right)$
graph{4x^3-12x^2+9x [-1, 3, -1.74, 3.26]}

It appears from the graph that we have a maximum volume of $2 f {t}^{2}$ when $x = \frac{1}{2}$, and a minimum volume of $0$ when $x = \frac{3}{2}$, so let us examine this further:

At a critical point (max or min) the derivative, $\frac{\mathrm{dV}}{\mathrm{dx}}$ will vanish. Differentiating wrt $x$ we get:

$\frac{\mathrm{dV}}{\mathrm{dx}} = 12 {x}^{2} - 24 x + 9$

At a critical point we have:

$\frac{\mathrm{dV}}{\mathrm{dx}} = 0 \implies 12 {x}^{2} - 24 x + 9 = 0$
$\therefore 4 {x}^{2} - 6 x + 3 = 0$
$\therefore \left(2 x - 1\right) \left(2 x - 3\right) = 0$

$x = \frac{1}{2} , \frac{3}{2}$

This is completely consistent with our graph so now let us prove the nature of the turning points by looking at the second derivative:

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

So:

$x = \frac{1}{2} \implies \frac{{d}^{2} V}{{\mathrm{dx}}^{2}} < 0 \implies$ maximum
$x = \frac{3}{2} \implies \frac{{d}^{2} V}{{\mathrm{dx}}^{2}} > 0 \implies$ minimum

Ths we get a maximum volume when:

$x = \frac{1}{2} \implies V = \left(\frac{1}{2}\right) {\left(3 - 2 \left(\frac{1}{2}\right)\right)}^{2} = 2$

Which matches are graphical observations.

Thus the maximum occurs when:

$x = \frac{1}{2} \implies y = 2$

$V = 2 \setminus f {t}^{3}$