Originally Posted by

**mellowdano** The question is : Find the dimensions of the rectangular box with largest volumes if the total surface area is given as 64cm^2. (it is a closed top box)

What I have done

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.

Is this correct or was i suppose to use a different equation?