# minimising volume

• Dec 10th 2008, 01:18 PM
Matho
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.
• Dec 10th 2008, 01:53 PM
skeeter
Quote:

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.
• Dec 10th 2008, 02:35 PM
Matho
Quote:

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.

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