Page 352 - DMTH201_Basic Mathematics-1

Unit 14: Business Applications of Maxima and Minima

For maximum TR, we have Notes
d TR = 34 - 0.4x = 0 or x 34 85 dozens.
dx 0.4
2
d TR
Since 2 = 4 0, the second order condition is satisfied.
dx
Alternate Method

Let y be number of dozens ordered in excess of 50 dozens.
Then p = 24 0.20y and quantity ordered = 50 + y.
TR = (24 0.20 ) (50y ) y

= 1200 + 14y – 0.20y 2
( d TR )
dy = 14 0.40y 0 for maxima.
14
or y = 35
0.40
d 2 (TR )
= –0.40 0
dy
Thus, revenue is maximised when (50 + 35) = 85 dozens of pencils are ordered.

Example: A tour operator charges 200 per passenger for 50 passengers with a discount
of 5 for each 10 passenger in excess of 50. Determine the number of passengers that will
maximise the revenue of the operator.
Solution:
Let x be the number of passengers, then revenue from each passenger i.e. price p is given by
5 x
p = 200 (x 50) 225
10 2

5
The equation of a straight line passing through the point (50, 200) with slope = .
10
x x 2
225 x 225x .
TR =
2 2
( d TR ) = 225 x
dx 0 or
d 2 (TR )
= 1 0
dx 2
TR is maximised when x = 225 passengers. Alternatively, we can write the revenue function
as.
y
TR 200 50 , y where y is the number of passengers in excess of 50.
2

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