A company wishes to manufacture a box w/ a volume of 20 cubic ft 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. Round to the nearest tenth, if necessary.

Question

anonymous · Answer

We want to minimize

M= 2lh +2wh +lw

Subject to

lwh= 20

and 2w=l

Substituting 2w for l in the first equation.

M= 4wh + 2wh + 2w^2= 2w^2 +6wh

Substituting 2w for l in lwh= 20

2w^2h=20
So h = 10/w^2

Substituting for h in M
M= 2w^2* 6w*10/w^2=2w^2+60/w

Taking the derivative we have
dM/dw= 4w -60/w^2

Setting it equal to 0
4w-60/w^2= 0
w^3=15
w=(15)^(1/3)
or 
approximately 2.47 feet. 

Make sense?