Page 340 - DMTH201_Basic Mathematics-1

Unit 13: Maxima and Minima

Notes
Example: Find the nature of point of inflexion of the following functions:
2
3
(i) y = x – 15x + 20x + 10
2
(ii) y = 20 + 5x + 12x – 2x 3
Solution:
dy
2
2
3
(i) y = x – 15x + 20x + 10 = 3x – 30x + 20
dx
2
d y
and = 6x – 30 = 0, for the point of inflexion x = 5
dx 2
3
d y
Further, 6, which is positive for all values of x.
dx 3
The point of inflexion at x = 5 is of type II i.e, curve changes from concave to convex
from below.
dy
(ii) y = 20 + 5x + 12x – 2x 3 = 5 + 24x – 6x 2
2
dx
2
d y
and = 24 – 12x = 0, for the point of inflexion x = 2
dx 2
3
d y
Further, = – 12 < 0 The point of inflexion at x = 2 is of type I i.e. the curve changes from
dx 3
convex to concave from below.
Example: Find maxima, minima and the points of inflexion for the following functions
and hence trace their curves:
3
(i) y = x + 10x + 25x – 40 (ii) y = x – 6x + 1
2
4
2
Solution:
(i) y = x + 10x + 25x – 40
3
2
First order condition ( maxima or minima)
dy
2
= 3x + 20x + 25 = 0 for maxima or minima or (3x + 5) (x + 5) = 0
dx
5
Thus, the stationary points are x and x = –5.
3
Second order condition
2
d y 5
= 6x + 20 = 10 > 0, when x .
dx 2 3
5
Thus, the function has a minima at x = –1.67.
3
The minimum value of the function f(–1.67) = –58.52.
2
d y
When x = –5 we have 2 = –30 + 20 = –10 < 0. Hence, the function has a maxima at
dx
x = –5. The maximum value of the function f(–5) = –40.

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