# What are the dimensions of a box that will use the minimum amount of materials, if the firm needs a closed box in which the bottom is in the shape of a rectangle, where the length being twice as long as the width and the box must hold 9000 cubic inches of material?

Feb 24, 2015

Let's begin by putting in some definitions.

If we call $h$ the height of the box and $x$ the smaller sides (so the larger sides are $2 x$, we can say that volume

$V = 2 x \cdot x \cdot h = 2 {x}^{2} \cdot h = 9000$ from which we extract $h$

$h = \frac{9000}{2 {x}^{2}} = \frac{4500}{x} ^ 2$

Now for the surfaces (=material)

Top&bottom: $2 x \cdot x$ times $2 \to$ Area=$4 {x}^{2}$
Short sides: $x \cdot h$ times $2 \to$ Area=$2 x h$
Long sides: $2 x \cdot h$ times $2 \to$ Area=$4 x h$

Total area:

$A = 4 {x}^{2} + 6 x h$

Substituting for $h$

$A = 4 {x}^{2} + 6 x \cdot \frac{4500}{x} ^ 2 = 4 {x}^{2} + \frac{27000}{x} = 4 {x}^{2} + 27000 {x}^{-} 1$

To find the minimum, we differentiate and set $A '$ to $0$

$A ' = 8 x - 27000 {x}^{-} 2 = 8 x - \frac{27000}{x} ^ 2 = 0$

Which leads to $8 {x}^{3} = 27000 \to {x}^{3} = 3375 \to x = 15$

Short side is $15$ inches
Long side is $2 \cdot 15 = 30$ inches
Height is $\frac{4500}{15} ^ 2 = 20$ inches
Check your answer! $15 \cdot 30 \cdot 20 = 9000$