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Daily production cost and profit maximization

CAT 2007 Slot 1 · Quantitative Ability · Hard · Functions

This is a hard Quantitative Ability question set from the CAT 2007 Slot 1 paper. It tests Functions. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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Passage / data set

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 'xx' units is 240+bx+cx2240 + bx + cx^2, where 'bb' and 'cc' are some constants. Mr. David noticed that doubling the daily production from 20 to 40 units increases the daily production cost by 6623%66\frac{2}{3}\%. However, an increase in daily production from 40 to 60 units results in an increase of only 50%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 1 of 2

How many units should Mr. David produce daily?

  1. A.

    130

  2. B.

    100

  3. C.

    70

  4. D.

    150

  5. E.

    Cannot be determined

Answer

B

Explanation

Cost function C(x)=240+bx+cx2C(x) = 240 + bx + cx^2. C(40)=53C(20)C(40) = \frac{5}{3} C(20) and C(60)=32C(40)C(60) = \frac{3}{2} C(40). Solving these gives b=10b = 10 and c=110c = \frac{1}{10}. Profit P(x)=30x−(240+10x+x210)=20x−240−x210P(x) = 30x - (240 + 10x + \frac{x^2}{10}) = 20x - 240 - \frac{x^2}{10}. To maximize profit, P′(x)=20−x5=0  ⟹  x=100P'(x) = 20 - \frac{x}{5} = 0 \implies x = 100.

Question 2 of 2

What is the maximum daily profit, in rupees, that Mr. David can realize from his business?

  1. A.

    620

  2. B.

    920

  3. C.

    840

  4. D.

    760

  5. E.

    Cannot be determined

Answer

D

Explanation

Maximum profit P(100)=20(100)−240−100210=2000−240−1000=760P(100) = 20(100) - 240 - \frac{100^2}{10} = 2000 - 240 - 1000 = 760.

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