# Question 05570

Dec 2, 2016

$x = \frac{1}{3} \left(8 - \sqrt{19}\right) = 1.214$ Corresponds to the maximum volume.

$\therefore x = 1.2$ inches

The associated max volume is given by:

$\setminus \setminus \setminus \setminus \setminus V = 4 {x}^{3} - 32 {x}^{2} + 60 x$
$\therefore V = 32.8$ 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 6" by 10", hence, wlog taking $w = 6$ we get:

Width: $x + w + x = 6 \setminus \setminus \implies w = 6 - 2 x = 2 \left(3 - x\right)$
Length: $x + l + x = 10 \implies l = 10 - 2 x = 2 \left(5 - x\right)$

Then the volume is given by:

$\setminus \setminus \setminus \setminus \setminus V = w l x$
$\therefore V = 2 \left(3 - x\right) 2 \left(5 - x\right) x$
$\therefore V = 4 x \left({x}^{2} - 8 x + 15\right)$
$\therefore V = 4 {x}^{3} - 32 {x}^{2} + 60 x$

Differentiating wrt $x$ gives us;

$\therefore \frac{\mathrm{dV}}{\mathrm{dx}} = 12 {x}^{2} - 64 x + 60$

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

$\therefore 12 {x}^{2} - 64 x + 60 = 0$
$\therefore 3 {x}^{2} - 16 x + 15 = 0$

To solve this quadratic I will complete the square:

${x}^{2} - \frac{16}{3} x + 5 = 0$
$\therefore {\left(x - \frac{16}{6}\right)}^{2} - {\left(\frac{16}{6}\right)}^{2} + 5 = 0$
$\therefore {\left(x - \frac{16}{6}\right)}^{2} = \frac{256}{36} - 5$
$\therefore {\left(x - \frac{16}{6}\right)}^{2} = \frac{19}{9}$
$\therefore x - \frac{16}{6} = \pm \frac{\sqrt{19}}{3}$
$\therefore x = \frac{16}{6} \pm \frac{\sqrt{19}}{3}$
$\therefore x = \frac{1}{3} \left(8 \pm \sqrt{19}\right)$
$\therefore x = 1.214 , 4.120$ (3dp)

We should check while value leads to a maximum volume

$\therefore \frac{{d}^{2} V}{\mathrm{dx}} ^ 2 = 24 x - 64$

$x = \frac{1}{3} \left(8 - \sqrt{19}\right) \implies \frac{{d}^{2} V}{\mathrm{dx}} ^ 2 < 0 \implies \max$
$x = \frac{1}{3} \left(8 + \sqrt{19}\right) \implies \frac{{d}^{2} V}{\mathrm{dx}} ^ 2 > 0 \implies \min$

graph{4x^3-32x^2+60x [-5, 5, -30, 40]}

Hopefully you can visually confirm the above