Page 362 - DMTH201_Basic Mathematics-1

Unit 14: Business Applications of Maxima and Minima

1 3 2 50 Notes
= x 3x x 300
10 3
For max. p, we should have x 0 and x 0

3 2 50
Now, x = x 6x 0 or 9x 2 180x 500 0
10 3

180 180 2 4 9 500 180 120
x =
18 18
300 50 60 10
Thus, x = or and x or
1 18 3 2 18 3
6 6 50 50
x
Further, x = x 6 . 6 4 0 at 1
10 10 3 3
6 10 10
= . 6 4 0 at x 2
10 3 3
50 2
Therefore, profit maximising output of the plant = or 16 .
3 3
1 50 2 50 2 50 2
Max. profit = 3 300 207.41
10 3 3 3
Thus the firm is incurring loss of 207.41. Since this loss is less than 300 (fixed cost), the firm
will continue production.

1
Example: If the total cost of a firm is C x 3 5x 2 30x 10, where C is the total cost and
3
x is the level of output, and price under perfect competition is given as 6, find for what value(s)
of x the profit will be maximised? Also find the value of maximum profit and comment on the
result.
Solution:

We can write
1 1
(x) = 6x x 3 5x 2 30x 10 x 3 5x 2 24x 10
3 3
We have x = x 2 10x 24 0 or x 2 10x 24 0 , for max. p

x 6 x 4 = 0 x 1 6 and x 2 4

Further, x = 2x 10 12 10 2 0, when x = 6
and = 8 10 2 0 , when x = 4

Thus, the profit is maximum when x = 6 units.
1 3 2
Maximum profit = 6 5 6 24 6 10 46 i.e. loss of 46.
3
Since this loss is greater than the loss of 10, when nothing is produced, the firm will discontinue
production.

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