### CAT 2007 Question Paper Question 11

Instructions

Directions for the following two questions:

Mr. David manufactures and sells a single product at a fixed price in a niche market. The selling price of each unit is Rs. 30. On the other hand, the cost, in rupees, of producing x units is $$240 + bx + cx^2$$ , where b and c are some constants. Mr. David noticed that doubling the daily production from 20 to 40 units increases the daily production cost by 66.67%. However, an increase in daily production from 40 to 60 units results in an increase of only 50% in the daily production cost. Assume that demand is unlimited and that Mr. David can sell as much as he can produce. His objective is to maximize the profit.

Question 11

# How many units should Mr. David produce daily?

Solution

Cost of 20 units = 240+20b+400c

Cost of 40 units = 240+40b+1600c = 5/3 * (240+20b+400c) => 720+120b+4800c = 1200+100b+2000c

=> 480 = 20b + 2800c => 120 = 5b + 700c

Cost of 60 units = 240+60b+3600c = 3/2 (240+40b+1600c) => 480 + 120b + 7200c = 720 + 120b + 4800c

=> 240 = 2400c => c = 1/10 and b = 10

Let the number of items needed for max profit be k

CP = $$240+10k+k^2/10$$

SP = 30k

Profit = SP - CP = $$30k - 240 - 10k - k^2/10$$ = $$20k - 240 - k^2/10$$

or Profit = $$\frac{1}{10} (-k^2 + 200k - 2400)$$

or, Profit = $$\frac{1}{10} (-(k^2 - 200k + 2400))$$

or, Profit = $$\frac{1}{10} (-(k^2 - 200k + 2400 + 7600 - 7600))$$

or, Profit = $$\frac{1}{10} (-(k^2 - 200k + 10000) + 7600)$$

or, Profit = $$\frac{1}{10} (-(k - 100)^2 + 7600)$$

To maximise the value of Profit, $$-(k - 100)^2$$ must be 0.

So, $$k$$ must be equal to 100.

Hence, option B is the correct answer.