Question

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

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

Let x, y, and z be three natural numbers

Then according to the question,

x + y + z = 24

Let's write z in terms of x and y, then

z = 24 - x- y

Now, let u be the function of x and y such that the product of the first, square of second, and cube of the third is maximum then,

$u = (24-x-y) x^{2}y^{3}$

$u = 24x^{2}y^{3} -x^{3}y^{3} - x^{2}y^{4}$              equation 1

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

$\frac{du}{dx} = 0\\\\48xy^{3} - 3x^{3}y^{3} - 2xy^{4} = 0\\xy^{3}(48 - 3x - 2y) = 0\\48 - 3x - 2y = 0\\3x + 2y = 48$              equation 2

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

$\frac{du}{dy} = 0\\\\72x^{2}y^{2} - 3x^{3}y^{2} - 4x^{2}y^{3} = 0\\x^{2}y^{2}(72 - 3x - 4y) = 0\\72 - 3x - 4y = 0\\3x + 4y = 72$          equation 3

Subtracting 2 from 3

$3x + 4y - 3x - 2y = 72 - 48$

$2y = 24$

∴y = 12

⇒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 $\frac{d^{2}u}{dx^{2}}$ and $\frac{d^{2}u}{dy^{2}}$  both should be less than zero.

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

$\frac{d^{2}u}{dx^{2}} = 48y^{3} - 6xy^{3} - 2y^{4}$

On substitution,

∴ $\frac{d^{2}u}{dx^{2}} = 48*1728 - 6*8*1728-2*12^{4} = 0-2*12^{4}\\\\\frac{d^{2}u}{dx^{2}} > 0$

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

$\frac{d^{2}u}{dy^{2}} = 144x^{2}y-6*3y-12*2y^{2} \\= 144*64*12 - 6*512*12-12*64*144\\= 0-6*512*12\\\frac{d^{2}u}{dy^{2}} < 0$

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

∴ The required numbers are 8, 12 and 4