The demand function is $p = 40 - 2x$ (price in ₹ when $x$ units are sold). Find the number of units that maximises the total revenue, and the maximum revenue.

The demand function is p = 40 - 2x (price in ₹ whe – Class 12 Mathematics And Statistics

Exercises · Q12

p = 40 - 2x

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Total revenue.

R(x) = p\,x = (40 - 2x)x = 40x - 2x^2.

Maximise. Marginal revenue:

\text{MR} = R'(x) = 40 - 4x.

Set $R'(x) = 0$ :

40 - 4x = 0 \ \Rightarrow\ x = 10.

Confirm a maximum. $R''(x) = -4 < 0$ , so $x = 10$ gives a maximum.

Maximum revenue.

R(10) = 40(10) - 2(10)^2 = 400 - 200 = 200.

So the maximum revenue is ₹200 (at a price $p = 40 - 2(10) = ₹20$ per unit). …