Question

Divide the number 48 into three parts such that its product is maximum.

Answer 1

Let x, y, z be the three parts of the number 48

∴ x + y + z = 48

Also, let u = xyz and

F = xyz + λ(x + y + z)

We form the equations F_x = 0, F_y = 0, F_z = 0

i.e., yz + λ = 0; xz + λ = 0; xy + λ = 0

or λ = – yz; λ = – xz and λ = – xy

⇒ – yz = – xz = – xy

and hence x = y = z

Since x + y + z = 48, we get

x = y = z = 16

Thus, 16, 16, 16 are the three parts of 48 such that the product is maximum.