Question

Divide 24 into 3 parts such that the product of the first, square of second and cube of the third is maximum.

Answer 1

Let x and y be two natural numbers. Let z be the third number such that x + y + z = 24 and z = 24 − x − y

Let u be the function of x and y such that the product of the first, square of second and cube of third is maximum.

Diff. u partially w.r.t.x and equating it to zero,

Diff u partially w.r.t.y and equating it to zero

Subtracting equation (iii) from equation (iv)

The product should be maximum, i.e. the function uu should have maximum value. For this condition to be satisfied, it’s necessary that ∂²u/∂x² and ∂²u/∂y² both should be less than zero.

Diff. equation (i) partially w.r.t. x,

On substitution,

Diff. equation (iii) partially w.r.t. y,

Both conditions are satisfying. Hence these numbers give maximum product.

∴ These numbers are 8, 12 and 4

Required numbers are 8, 12 and 4.