Page 158 - DECO403_MATHEMATICS_FOR_ECONOMISTS_ENGLISH

Unit 9: Maxima and Minima: One Variable

Note
2
dy 1
= – ey is negative, therefore for the function x = is maximum
dx 2 e
1
Thus, in the referred function replacing with x =
e

1/e
§ 1 ·
¨ 1 ¸
1/e
Maximum value = ¨ ¸ = (e) is proven.
© e ¹
log x
Example 8: Find out the maximum value of where 0 < x < ∝∝ ∝∝ ∝
x

log x
Solution: Assume that y =
x
dy x .(1 / ) (log ).1 1log x
x
x

dx = x 2 = x 2
2
dy = x 2 (1 / ) (1 log ).2x
x
x

dx 2 x 4
x
x 2x 2 log x 2 log x 3
=
x 4 x 3
dy 1log x

Putting , =0 , = 0 or 1 – log x = 0 or log x = 1 = log e, ∴ x = e
dx x 2
2
dy 2log e 3 2 3 1
At x = e = 3 is negative
dx 2 e 3 e 3 e
log e 1
Therefore, at x = +e, function is maximum and its maximum value is = .
e e
Example 9: If at extremum values of x = –1 and x = 2 are y = a log x + bx + x then find out the value
2
of a and b.
dy 1
Solution: y = f(x) = a log x + bx + x ⇒ = a. + 2bx + 1
2
dx x
dy
For extremum = 0,
dx
§ dy · § dy ·
¨ ¸ = 0, ¨ dx ¸ = 0
© dx ¹ 1 © 2 ¹
⇒ – a – 2b + 1 = 0 ...(i)
a
§·
Or ¨¸ + 4b + 1 = 0 ...(ii)
2
©¹

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