# If 2400 square centimeters of material is available to make a box with a square base and an open top, how do you find the largest possible volume of the box?

Nov 3, 2015

The dimension of the box I found are:
$28.3 \times 28.3 \times 14.1 c m$ for a volume of $11 , 292.5 c {m}^{3}$:

#### Explanation:

The surface $S$ of a box with open top is the sum of the surfaces of a square (base, of side $a$) and 4 rectangles (base $a$ and height $h$):
$S = {a}^{2} + 4 a \cdot h$ (1)
while the volume $V$ will be the area of the base times the height, or:
$V = {a}^{2} \cdot h$ (2)

We get $h = \frac{S - {a}^{2}}{4 a}$ from (1) put it into (2):

$V = {a}^{2} \frac{S - {a}^{2}}{4 a} = \frac{1}{4 a} \left(S {a}^{2} - {a}^{4}\right) = \frac{S}{4} a - {a}^{3} / 4$
maximize this volume deriving with respect to $a$ and setting it equal to zero:
$\frac{\mathrm{dV}}{\mathrm{da}} = \frac{S}{4} - \frac{3}{4} {a}^{2} = 0$
$\frac{3}{4} {a}^{2} = \frac{S}{4}$
${a}^{2} = \frac{S}{3}$
$a = \pm \sqrt{\frac{2400}{3}} = \pm 28.3 c m$

We use $a = + 28.3 c m$ that in (1) gives us:
$h = \frac{2400 - {28.3}^{2}}{4 \cdot 28.3} = 14.1 c m$

Giving a volume of:
$V = {a}^{2} \cdot h = 11 , 292.5 c {m}^{3}$