If 1200 cm^2 of material is available to make a box with a square base and open top, find the largest possible volume of the box.

For maximum volume
$x = 2y$
 

 

$A = x^2 + 4xy$

$1200 = (2y)^2 + 4(2y)(y)$

$12y^2 = 1200$

$y = 10 ~ \text{cm}$
 

$x = 2(10) = 20 ~ \text{cm}$
 

$V_{max} = x^2 y = 20^2 (10)$

$V_{max} = 4000 ~ \text{cm}^3$

Available materialto make the box
$A = x^2 + 4xy$

$1200 = x^2 + 4xy$

$y = \dfrac{1200 - x^2}{4x}$
 

 

Volume of the box
$V = x^2 y$

$V = x^2 \left( \dfrac{1200 - x^2}{4x} \right)$

$V = 300x - \frac{1}{4}x^3$
 

For the largest possible volume of the box
$\dfrac{dV}{dx} = 300 - \frac{3}{4}x^2 = 0$

$300 = \frac{3}{4}x^2$

$x^2 = 400$

$x = 20 ~ \text{cm}$
 

Hence,
$V_{max} = 300(20) - \frac{1}{4}(20^3)$

$V_{max} = 4000 ~ \text{m}^3$