Question

divide 24 in 3 parts such that the continued product of the first, square of second and cube of the third is maximum using Lagrange's method.​

Answer 1

Answer:

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

Let uu be the function of xx and yy such that the product of the first, square of second and cube of third is maximum.

∴u=(24−x−y)x2y3u=24x2y3−x3y3−x2y4∴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)∂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)∂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=82y=24y=12∴x=48−243=8∴x=8

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∂2u∂x2 and ∂2u∂y2∂2u∂y2 both should be less than zero.

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

∴∂2u∂x2=48y3−6xy3−2y4∴∂2u∂x2=48y3−6xy3−2y4

On substitution,

∂2u∂x2=48×1728−6×8×1728−2×124=0−2×124∴∂2u∂x2<0∂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∴∂2u∂y2=144x2y−6x3y−12x2y2=144×64×12−6×512×12−12×64×144=0−6×512×12∴∂2u∂y2<0

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

∴∴ These numbers are 8, 12 and 4

Required numbers are 8, 12 and 4