Question

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

Answer 1

Answer:

These numbers are 8, 12 and 4

Step-by-step explanation:

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.

∴u=(24−x−y)x2y3u=24x2y3−x3y3−x2y4

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

∂u∂x=0∴48xy3−3x3y3−2xy4=0From(i)∴xy3(48−3x−2y)=0∴48−3x−2y=0[∵x≠0,y≠0]∴3x+2y=48…(ii)

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

∂u∂y=0∴72x2y2−3x3y2−4x2y3=0…(iii)∴x2y2(72−3x−4y)=0∴72−3x−4y=0[∵x≠0,y≠0]∴3x+4y=72…(iv)

Subtracting equation (iii) from equation (iv)

2y=24y=12∴x=48−243=8∴x=8

The product should be maximum, i.e. the function u 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,

∴∂2u∂x2=48y3−6xy3−2y4

On substitution,  

∂2u∂x2=48×1728−6×8×1728−2×124=0−2×124∴∂2u∂x2<0

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

∴∂2u∂y2=144x2y−6x3y−12x2y2=144×64×12−6×512×12−12×64×144=0−6×512×12∴∂2u∂y2<0