1. ## minimising volume

A company wishes to manufacture a box with a volume of 32 cubic feet that is open on top and is twice as long as it is wide. Find the width of the box that can be produced using the minimum amount of material.

2. Originally Posted by Matho
A company wishes to manufacture a box with a volume of 32 cubic feet that is open on top and is twice as long as it is wide. Find the width of the box that can be produced using the minimum amount of material.
$V = L \cdot W \cdot H$

$L = 2W$

$32 = 2W \cdot W \cdot H$

$\frac{16}{W^2} = H$

surface area ...

$A = (2W \cdot W) + 2(W \cdot H) + 2(2W \cdot H)$

get $A$ in terms of a single variable ( $W$), find $\frac{dA}{dW}$, set it equal to 0, and find the value of $W$ that minimizes the area.

3. Originally Posted by skeeter
$V = L \cdot W \cdot H$

$L = 2W$

$32 = 2W \cdot W \cdot H$

$\frac{16}{W^2} = H$

surface area ...

$A = (2W \cdot W) + 2(W \cdot H) + 2(2W \cdot H)$

get $A$ in terms of a single variable ( $W$), find $\frac{dA}{dW}$, set it equal to 0, and find the value of $W$ that minimizes the area.

4. $4W - \frac{96}{W^2} = 0$
$4W = \frac{96}{W^2}$
$W^3 = 24$
$W = 2\sqrt[3]{3}$