minimising volume

**Matho** · December 10th 2008, 01:18 PM

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.

**skeeter** · December 10th 2008, 01:53 PM

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.

**Matho** · December 10th 2008, 02:35 PM

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.

Thanks for replying me.
I got da/dW' = 4w - 96/w^2 but how do I get equal 0?