Paper 4: Fundamentals of Business Mathematics & Statistic

3.48 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS

Calculus

Find

2

2

d y.

dx Put x = a in

2

2

d y,

dx if

2

2

d y

dx at x = a is – ve, the function is maximum at x = a and maximum value is

f(a).

If again by putting x = a in

2

2

d y,

dx the result is +ve, then the function is minimum and minimum value is f(α).

Similarly for the value x = b.

Example 104 : Examine maximum value of the function

y = x^3 – 27x + 10 (the same example given above)

Solution:

dy 3x 27. Taking 0 i.e., 3x 27 0, 2 dy 2

dx= − dx= − = we get x = ± 3.

Now

2

2

d y 6x.

dx = At x = 3,

2

2

dy 6.

dx = 3 = 18, + ve, so the function is minimum at x = 3 and min. value is

33 – 27.3 + 10 = – 44.

Again at x ( )

2

2

3,d y 6. 3 18, ve,

= − dx = − = − − so the function is max. at x = – 3 and max. value is

(– 3)^3 – 27 (– 3) + 10 = 64.

Example 105 :For what value of x the following function is maximum or minimum,

x 7x 6^2

x 10−− +.

(Use of Second Derivative Method)

Let y =

x 7x 6^2

x 10

− +

−

dy

dx=

( )( ) ( )

( ) ( )

(^22)
2 2
x 10 2x 7 x 7x 6 x 20x 64
x 10 x 10

− − − − + − +

=

− −

making dydx= 0 we find x^2 – 20x + 64 = 0

or, (x – 4) (x – 16) = 0 ∴ x = 4, 16.

Now

( ) ( ) ( ) ( )

( )

2 2 2

2 3

d y x 10 2x 20 x 20x 64 .2 x 10

dx x 10

− − − − + −

=

−

{( ) ( )( )}

( )

2

3

2 x 10 x 4 x 16

x 10

− − − −

=

−

At x = 4,

( )

( )

2 2

2 3

d y 2 4 10 2 1, ve

dx 4 10 16 3

= − = = − −

− −^ ∴ function is max. at x = 4.

At x = 16,

( )

( )

2 2

2 3

d y 2 16 10 2 1 ve

dx 16 10 6 3

= − = = +

−^ ∴ function is min. at x = 16.