An open rectangular box has volume 32 cm3. find the lengths of the edges giving the minimum surface area

Open box, so there is no top 
SA = 2*l*h + 2*w*h + l*w 

V = l*w*h = 32cm^3 
h = 32/(w*l) 

SA = 2*l*32/(w*l) + 2*w*32/(w*l) + l*w 
SA = 64/w + 64/l + l*w 
Find minimum SA: take partial derivatives to get critical point(s) 
SAw = -64/w^2 + l 
SAl = -64/l^2 + w 
Both the partials have to be 0, so... 
0 = -64/w^2 + l and 0 = -64/l^2 + w 
64/w^2 = l 
0 = -64/(64/w^2)^2 + w (plug into second equation) 
0 = -w^4/64 + w 
0 = w(1-w^3/64) 
1 = w^3/64 or 0 = w (impossible answer) 
64 = w^3 
4 = w 

Plug w back into 64/w^2 = l 
64/4^2 = l 
4 = l 

Plug w and l back into h = 32/(w*l) 
h = 32/(4*4) 
h = 2 

The Surface Area:2*l*h + 2*w*h + l*w 
SA = 2*4*2 + 2*4*2 + 4*4 
SA = 48 

So the answer is length = 4cm, width = 4cm, and height = 2cm, with Surface Area of 48cm^2