Question

An open top rectangular box must have a volume of 4500 cu ft. If the bottom is a rectangle with length is twice the width, what dimensions would minimize total surface area of the box?

Answer 1

V = l w h
SA = 2(lw + lh + wh)

l = 2w

V = 2 w w h
SA = 2(2w w + 2wh + wh)

4500 = 2 w w h
h = 4500 / 2 w w

SA = 2(2 w w + 2 w (4500 / 2 w w ) + w (4500 / 2 w w ))
SA = 4*w w+13500*w w w

SA' = ...
Minimize it simply