# An open top box is to be constructed so that its base is twice as long as it is wide. Its volume is to be 2400cm cubed. How do you find the dimensions that will minimize the amount of cardboard required?

Nov 22, 2015

The problem provides the secondary equation relating the sides to volume. Use that to derive a one-variable equation.

#### Explanation:

Secondary Equation : $w \times \left(2 w\right) \times h = 2400$

Solve for h:

$h = \frac{1200}{w} ^ 2$

Primary Equation for Surface Area (with open top) :
$f \left(w\right) = 2 {w}^{2} + 2 w h + 2 \left(2 w\right) h$

Now use the Secondary Equation by substituting for h:

$f \left(w\right) = 2 {w}^{2} + \left(2 w\right) \left(\frac{1200}{w} ^ 2\right) + \left(4 w\right) \left(\frac{1200}{w} ^ 2\right)$

Simplify:

$f \left(w\right) = 2 {w}^{2} + \frac{7200}{w}$

Now find the derivative and set equal to zero:

$f ' = 4 w - \frac{7200}{w} ^ 2 = 0$

${w}^{3} = 1800$

$w = {\left(1800\right)}^{\frac{1}{3}} \approx 12.1644$

[I'll leave it for you to prove that this is a minimum]

Finally, the dimensions that minimize the surface area are:

$w = {\left(1800\right)}^{\frac{1}{3}}$
$l = 2 {\left(1800\right)}^{\frac{1}{3}}$
$h = \frac{1200}{w} ^ 2 = \frac{1200}{1800} ^ \left(\frac{2}{3}\right)$

hope that helped