‹ Back to the paper
Solve the following.
The demand for a certain product is represented by the equation in rupees where is the number of…
The demand for a certain product is represented by the equation $p = 500 + 25x - \frac{x^2}{3}$ in rupees where $x$ is the number of units and $p$ is the price per unit. Find:
(i)[2.5]
Marginal revenue function.
(ii)[2.5]
The marginal revenue when 10 units are sold.
Answer
Answer (i)
AIRevenue $R = px = 500x + 25x^2 - \frac{x^3}{3}$.
Marginal revenue $MR = \frac{dR}{dx}$
Hence $MR = 500 + 50x - x^2$.
Final answer: $500 + 50x - x^2$
Answer (ii)
AIAt $x = 10$: $MR = 500 + 50(10) - 10^2 = 500 + 500 - 100$.
Hence the marginal revenue is ₹ 900.
Final answer: $900$ ₹
From ISC 2017 Mathematics Paper 1, question 14(a).