An open rectangular box is 4 feet long and has a surface area of 20 square feet. find the dimensions of the box for which the volume is as large as possible.

can someone explain the steps to the answer if you can?

SA=2H(L+W)+LW for open box
20=2H(4+W)+4(W)
distribute
20=8H+2WH+4W
divide both sides by 2
10=4H+WH+2W
solve for 1 variable, pick W
10-4H=WH+2W
10-4H=W(H+2)
(10-4H)/(H+2)=W

V=LWH
subsitute 4 for V, subsitute H for H and (10-4H)/(H+2) for W
V=(4)(H)(10-4H)/(H+2)
V=(40H-16H²)/(H+2)
find max value
take deritivitive of this thing
V'=-16(H²+4H-5)/((H+2)²)
using sign chart
sign changes from positive to negative at H=1

so at H=1
find W

W=(10-4H)/(H+2)
W=2

the dimeionts are

length=4ft
width=2ft
height=1ft
(the volume is 8 cubic feet)