The marginal cost (MC) and marginal revenue (MR) functions of a product are $MC = 20+\frac{x}{20}$ and $MR = 30$ respectively. If the fixed cost is 200, then the maximum value of the profit is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Rs. 800
Given:
$MC = 20 + \frac{x}{20}, \quad MR = 30, \quad \text{Fixed cost} = 200$
Total cost (TC):
$TC = \int MC \, dx + 200 = \int \left(20 + \frac{x}{20}\right) dx + 200$
$= 20x + \frac{x^2}{40} + 200$
Total revenue (TR):
$TR = \int MR \, dx = \int 30 \, dx = 30x$
Profit function:
$P(x) = TR - TC = 30x - \left(20x + \frac{x^2}{40} + 200\right)$
$= 10x - \frac{x^2}{40} - 200$
To maximize profit, differentiate:
$\frac{dP}{dx} = 10 - \frac{x}{20}$
Set $\frac{dP}{dx}=0 \;\;\Rightarrow\;\; 10 - \frac{x}{20} = 0 \;\;\Rightarrow\;\; x = 200$
Profit at $x=200$:
$P(200) = 10(200) - \frac{200^2}{40} - 200$
$= 2000 - 1000 - 200$
$= 800$
therefore, The maximum value of the profit is $800$.