Mumbai University > First Year Engineering > Sem 1 > Applied Maths 1

Marks : 6 M

Year : May 2014

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.

$ \\ \therefore u\;=\; (24-x-y) x^2 y^3 \\ \; \\ \; \\ u \;=\; 24x^2y^3 -x^3y^3 - x^2y^4 \\ \; \\ $

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

$ \dfrac{\partial u}{\partial x} \;=\; 0 \\ \; \\ \therefore 48xy^3-3x^3y^3-2xy^4 \;=\; 0 \;\;\; From \; (i) \\ \; \\ \therefore xy^3 (48-3x-2y) \;=\; 0 \therefore 48-3x-2y=0 \; \; [\because x \ne 0 \; ,\; y \ne 0] \\ \; \\ \therefore 3x+2y = 48 \; \; \; \ldots (ii) \\ \; \\ $

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

$ \dfrac{\partial u}{\partial y} \;=\; 0 \\ \; \\ \therefore 72x^2y^2 -3x^3y^2-4x^2y^3 \;=\; 0 \; \; \; \ldots (iii) \\ \; \\ \therefore x^2y^2 (72-3x-4y) \;=\; 0 \\ \; \\ \therefore 72-3x-4y=0 \; \; [ \because x \ne 0 \;, \; y \ne 0 ] \\ \; \\ \therefore 3x+4y=72 \; \; \; \ldots (iv) \\ \; \\ \; \\ $

Subtracting equation (iii) from equation (iv)

$ 2y \;=\; 24 \\ y \;=\; 12 \; \; \; \; \therefore x \;=\; \dfrac{48-24}{3} \;=\; 8 \; \; \; \; \therefore 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 $ \dfrac{\partial^2 u}{\partial x^2} $ and $ \dfrac{\partial^2 u}{\partial y^2} $ both should be less than zero.

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

$ \therefore \dfrac{\partial^2 u}{\partial x^2} \;=\; 48y^3 -6xy^3-2y^4 \\ \; \\ $

On substitution,
$ \dfrac{\partial^2 u}{\partial x^2} \;=\; 48 \times 1728 - 6 \times 8 \times 1728 - 2 \times 12^4 \;=\; 0 - 2 \times 12^4 \\ \; \\ \therefore \dfrac{\partial^2 u}{\partial x^2} \; \lt \; 0 \\ $

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

$ \therefore \dfrac{\partial^2 u}{\partial y^2} \;=\; 144x^2y -6x^3y-12x^2y^2 \\ \; \\ = 144 \times 64 \times 12 - 6 \times 512 \times 12 - 12 \times 64 \times 144 \\ \; \\ = 0-6 \times 512 \times 12 \\ \; \\ \; \\ \therefore \dfrac{\partial^2 u}{\partial y^2} \; \lt \; 0 \\ \; \\ $

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

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$ \therefore$ These numbers are 8, 12 and 4

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Required numbers are 8, 12 and 4