Skip to content

CA Foundation · Quantitative Aptitude · Differential and Integral Calculus

The demand function for a product is x = 60 − 3p, where x is the quantity demanded and p is the price in ₹. The price at which total revenue is maximum is:

Writing revenue as p(60 − 3p) = 60p − 3p², its derivative 60 − 6p equals zero at p = 10, and the second derivative is negative. So revenue is maximised at a price of ₹10.

  1. A₹10Correct
  2. B₹20
  3. C₹30
  4. D₹15

Explanation

Here p = (60 − x)/3 = 20 − x/3. Revenue R = px = 20x − x²/3. R' = 20 − 2x/3 = 0 gives x = 30, and R'' = −2/3 < 0, so it is a maximum. Price = 20 − 30/3 = ₹10. Check using R = p(60 − 3p) = 60p − 3p², whose derivative 60 − 6p = 0 gives p = 10. ₹20 is the price at zero quantity, not at maximum revenue.

Did you get it right without looking?

One question tells you little. A timed set on Differential and Integral Calculus shows your real accuracy, how long you take and where you lose marks.

Practise Differential and Integral CalculusFree diagnostic mock

More Differential and Integral Calculus questions