The length of a cedar chest is twice its width. the cost of the lid is four times the cost of the rest of the cedar chest. If the volume is 1440 dm^3, find the dimensions of the box so that the cost is a minimum.

V=LxWxH

1440=2x(x)H

H=2x^2/1440

Solve for total Cost:

_____[Lid]____[Base]___[Long Sides]____[Short Sides]___

c=4c(2x^2)+c(2x^2)+c(2x)(2x^2)/1440+c(x)(2x^2)/1440

Find the derivative:

c'=16cx+4cx-8cx(1440^-2)-4cx(1440^-2)

Let c' = 0 to solve for x:

This again is where it get lost, letting the function = 0 is an ongoing problem for me.