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:

Rs. 800

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 $800$.