# Squares are cut from each corner of a 0.5m xx 0.5m square of cardboard and the sides folded up to make an open box. What is the maximum possible volume of the box?

May 29, 2015

Consider the general case of a square sheet of cardboard with sides of length $s$ and square cut outs at each corner with side $t$.

The area of the bottom of the resulting box will be ${\left(s - 2 t\right)}^{2}$.
The height of the box will be $t$.
The volume of the box will be $a r e a \times h e i g h t = {\left(s - 2 t\right)}^{2} \cdot t$

${\left(s - 2 t\right)}^{2} \cdot t = \left({s}^{2} - 4 s t + 4 {t}^{2}\right) \cdot t$

$= 4 {t}^{3} - 4 s {t}^{2} + {s}^{2} t$

The maximum of this will occur at some point where the derivative is zero...

$0 = \frac{d}{\mathrm{dt}} \left(4 {t}^{3} - 4 s {t}^{2} + {s}^{2} t\right) = 12 {t}^{2} - 8 s t + {s}^{2}$

$= 12 {t}^{2} - 6 s t - 2 s t + {s}^{2}$

$= \left(12 {t}^{2} - 6 s t\right) - \left(2 s t - {s}^{2}\right)$

$= 6 t \left(2 t - s\right) - s \left(2 t - s\right)$

$= \left(6 t - s\right) \left(2 t - s\right)$

So $t = \frac{s}{6}$ or $t = \frac{s}{2}$

If $t = \frac{s}{2}$ the volume is $0$ because $\left(s - 2 t\right) = 0$

So the maximum volume occurs when $t = \frac{s}{6}$.

Then

$v o l u m e = {\left(s - 2 t\right)}^{2} \cdot t$

$= {\left(s - 2 \left(\frac{s}{6}\right)\right)}^{2} \cdot \frac{s}{6}$

$= {\left(\frac{2 s}{3}\right)}^{2} \cdot \frac{s}{6}$

$= \frac{4}{54} {s}^{3}$

$= \frac{2}{27} {s}^{3}$

If $s = 0.5 m$

${s}^{3} = 0.125 {m}^{3}$

and

$\frac{2}{27} {s}^{3} = \frac{1}{108} {m}^{3} = 0.00 \dot{9} 2 \dot{5} {m}^{3} = 9259. \dot{2} 5 \dot{9} c {m}^{3}$