Q.A company's demand function is $p = 80 - x$ (price in ₹ per unit) and its total cost is $C(x) = 100 + 30x$. Determine the output that maximises profit, the maximum profit, and the selling price.

Build profit $\pi = R - C$ as a function of $x$, maximise it, and convert the optimal output into a price.

Revenue. $R(x) = p\,x = (80 - x)x = 80x - x^2$.

Profit.

Maximise.

Second derivative: $\pi''(x) = -2 < 0$, so $x = 25$ gives a maximum.

Maximum profit.

Selling price. $p = 80 - x = 80 - 25 = 55$, i.e. ₹55 per unit. …

Alternatively, apply MR = MC directly: $80 - 2x = 30 \Rightarrow x = 25$, then su …

The most frequent mistakes are subtracting the cost incorrectly (combine $-x^2 + 80x - 30x = -x^2 + 50x$) and forgetting to find the price from the demand relation. …