CAT WhatsApp Group
Sign in
Please select an account to continue using cracku.in
↓ →
CAT WhatsApp Group
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.
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.
Click on the Email ☝️ to Watch the Video Solution
Create a FREE account and get:
