Show that a rectangular parallelepiped of maximum volume V with constant surface area S is a cube?

Surface Area Of A Parallelepiped

This problem is the same as fixing the volume and minimizing the surface area.

v=xyz is constant
s=2xy+2xz+2yz=2v(1/x+1/y+1/z)
let p=1/x, q=1/y, r=1/z
s=2v(p+q+r)

s is a minimum when (p+q+r) is a minimum. pqr is also a constant. The sum will be a minimum when p=q=r. This is the same as when x=y=z.

x: length of the rectangular parallelepiped

y: width of the rectangular parallelepiped

z: height of the rectangular parallelepiped



The volume of the rectangular parallelepiped is:

V = xyz ← this is a fixed value



The surface area of the rectangular parallelepiped is:

S = (2 * yz) + (2 * xz) + (2 * xy) → recall: V = xyz → z = V/xy

S = [2 * y * V/(xy)] + [2 * x * V/(xy)] + (2 * xy)

S = [2 * V/x] + [2 * V/y] + (2 * xy)

S = (2V/x) + (2V/y) + 2xy → you can imagine that: y = k.x → where k is a constant

S = (2V/x) + (2V/kx) + 2kx² ← you can see that the volume is a function of x

…and you can get the minimum of a function when its derivative is null


S = (2V/x) + (2V/kx) + 2kx² ← this is the function where V is a given, i.e. a fixed value

S = [2V * (1/x)] + [(2V/k) * (1/x)] + 2kx² → recall the derivative of (1/x) is (- 1/x²)

S' = [2V * (- 1/x²)] + [(2V/k) * (- 1/x²)] + 4kx


Then you solve the equation: S' = 0

[2V * (- 1/x²)] + [(2V/k) * (- 1/x²)] + 4kx = 0

[2V * (- 1/x²)] + [(2V/k) * (- 1/x²)] = - 4kx

(- 1/x²).[2V + (2V/k)] = - 4kx

(1/x²).[2V + (2V/k)] = 4kx

2V + (2V/k) = 4kx³

V + (V/k) = 2kx³

x³ = [V + (V/k)] / 2k

x³ = (Vk + V)/2k² → suppose that: x = y → i.e. the base of the rectangular parallelepiped is a square


You can deduce that: k = 1, because recall: y = k.x


x³ = (Vk + V)/2k² → where: k = 1

x³ = (V + V)/2

x³ = 2V/2

x³ = V ← this result is possible only if the rectangular parallelepiped is a cube where x is the edge


As we've just seen that x = y

V = xyz

z = V/xy

z = x³/xy

z = x²/y → recall: x = y

z = y²/y

z = y


You can conclude that: x = y = z → for a given volume, the surface area will be minimized when the volume is a cube.