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.