Optimizing a minimum

Question

A closed rectangular container with a square base is to have a volume of 3456 in3. The material for the top and bottom of the container will cost per $2 in^2, and the material for the sides will cost per $1 in^2. Find the dimensions of the container of least cost.

Answer 1

V=LWH = 3456

L=W

C=2LW*(2)+2H(L+W)(1)

substitute W for L and find C=4W²+4WH

Find dC/dW=8W+4WdH/dW+4H

Set dC/dW=0 finding dH/dW= -(2W+H)/W

Note: V=LWH=HW²=3456

Find dV/dW=2HW+w²dH/dW =0

solve for dH/dW=-2H/W³

substitue -2H/W3=-(2W+H)/W and H=3456/W²

solve to find W=12 and L=12 and H=24