the length of a cedar chest is twice its width. the cost/dm^2 of the lid is four times the cost dm^2 of the rest of the cedar chest. if the volume of the chest is 1440 dm^3 find the dimensions so that the cost is a minimum.

this is what i got so far

V=L x W x H

1440=2x(x)h

1440=2x^2(h)

h=720/x^2

C=4c(area of lid) + c(area of base) + c(area of sides)

That is what i got for cost and for the derivative, i'm having trouble solving for x and i should be able to get the dimensions once i have the value of x