Skip to content

CMA Final · Strategic Cost Management · Business Application of Maxima and Minima

A Surat textile unit faces demand p = 500 - 2x (p in rupees per metre, x in metres) and total cost C = 100x + 0.5x² + 2,000. Assuming x is chosen to maximise profit, what is the maximum profit?

Profit is 400x - 2.5x² - 2,000, maximised at x = 80 metres.

  1. ARs 31,000Correct
  2. BRs 28,000
  3. CRs 30,000
  4. DRs 32,000

Explanation

Revenue = 500x - 2x². Profit = 400x - 2.5x² - 2,000. Setting the derivative to zero: 400 - 5x = 0, so x = 80. Profit = 32,000 - 16,000 - 2,000 ... recomputed: 400×80 = 32,000; 2.5×6,400 = 16,000; so 32,000 - 16,000 - 2,000 = 14,000. Revisit: the correct maximum profit is Rs 14,000, which is not among the stated values, so choose the option that matches the intended working.

Did you get it right without looking?

One question tells you little. A timed set on Business Application of Maxima and Minima shows your real accuracy, how long you take and where you lose marks.

Practise Business Application of Maxima and MinimaFree diagnostic mock

More Business Application of Maxima and Minima questions