Paper 4: Fundamentals of Business Mathematics & Statistic

3.44 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS

Calculus

LAGRANGES METHOD MULTIPLIERS :

To find the extreme of a differentiable function f(x, y) of two variables subject to the condition of independent

and differentiable equation g (x, y) = 0

Here f(x, y) depends in reality on only two independent variables x and y. For symmetry multiply

f(x, y) by 1 and equ. (i) by l (a constant) and add them together so that we get,

L = 1. f (x, y) + l g (x, y)

For maximum or minimum

L L 0

x y

∂ =∂ =

∂ ∂ ... (ii)

Now solving equ. (i) and (ii) we may find the values of z and y, and hence the extremes.

Example 100 : Find the extreme value of the function

f (x, y) = x^2 – y^2 + xy + 5x. Subject to x + y + 3 = 0

Let g (x, y) = x + y + 3

L = 1. f (x, y) + l g(x, y) = x^2 – y^2 + xy + 5x + l ( x + y + 3).

For max. or min.,

L L 0

x y

∂ =∂ =

∂ ∂

i.e., 2x + y + 5 + l = 0 ... (i), – 2y + x + l = 0 ...(ii)

Again x + y + 3 = 0 (given) ...(iii) Solving (i) (ii) (iii)

We get x = – 2, y = – 1, l = 0 so at ( – 2, – 1)

The given function has extreme value and the value is

f (– 2, – 1) = 4 – 1 + 2 – 5.2 = 6 – 11 = – 5.

Example 101: Find the extreme values of the function.

f (x, y) = x^2 – y^2 + xy + 5x subject to x – 2y = 0

Let L = 1 f(x, y) + lg (x, y), where g (x, y) = x – 2y; L = x^2 – y^2 + xy + 5x +l (x – 2y).

From max. or min.

L L 0

x y

∂ =∂ =

∂ ∂ so that

L

x

∂

∂ = 2x + y + 5 + l = 0 ... (i)

L 2y x 2 0

x

∂ = − + − λ =

∂ ... (ii)

In equ. (i), putting x = 2y as x – 2y = 0 we find 4y + y + 5 + l or, l = 0

From (iii), 5y + 5 = 0 or, y = – 1 as l = 0 ∴ x = 2y = 2 (–1) = – 2

At (– 2, – 1), extreme value = 4 – 1 + 2 + 5 (– 2) = – 5.

Example 102: Find the minimum value of f(x, y) = x^2 + y^2 subject to x + y = 10

Let g (x) = x + y = 10. L = 1. f(x) + lg (x) = x^2 + y^2 + l (x + y – 10).

For minimum value

L L 0 i.e.

x y

∂ =∂ =

∂ ∂ 2x + l = 0 .... (i) 2y + l = 0 ... (ii)