Find the dimensions of the rectangular box with largest volume if the total surface area is given as 64cm^2

Let x =lenght, y = width, and z =height 
The volume of the box is equal to V = xyz 
Subject to the surface area 
S = 2xy + 2xz + 2yz = 64 
= 2(xy + xz + yz) 
= 2[xy + x(64/xy) + y(64/xy)] 
S(x,y)= 2(xy + 64/y + 64/x) 
Then 
Mx(x, y) = y = 64/x^2 
My(x, y) = x = 64/y^2 
y^2 = 64/x 
(64/x^2)^2 = 64 
4096/x^4 = 64/x 
x^3 = 4096/64 
x^3 = 64 
x = 4 
y = 64/x^2 
y = 4 
z= 64/yx 
z= 64/16 
z = 4 

Therefor the dimensions are cube 4.