A company is selling a certain commodity $x$. The demand function for the commodity is linear. The company can sell $2000$ units when the price is ₹$8$ per unit and it can sell $3000$ units when the price is ₹$4$ per unit. The Marginal revenue at $x=5$ is:

The demand function is linear, so it has the form $p = mx + c$, where $p$ is the price per unit and $x$ is the number of units sold.

We're given two points on this line:

• When $x = 2000$, $p = 8$
• When $x = 3000$, $p = 4$

Finding the slope:

$m = \dfrac{4 - 8}{3000 - 2000} = \dfrac{-4}{1000} = -\dfrac{1}{250}$

Plugging $(2000, 8)$ into $p = mx + c$:

$8 = -\dfrac{1}{250}(2000) + c$

$8 = -8 + c$

$c = 16$

So the demand function is:

$p = 16 - \dfrac{x}{250}$

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Revenue $= p \cdot x$, so:

$R(x) = \left(16 - \dfrac{x}{250}\right) \cdot x$

$R(x) = 16x - \dfrac{x^2}{250}$

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Marginal Revenue is the derivative of $R(x)$ with respect to $x$. It tells us how much extra revenue we earn by selling one more unit.

$MR = \dfrac{dR}{dx} = \dfrac{d}{dx}\left(16x - \dfrac{x^2}{250}\right)$

$MR = 16 - \dfrac{2x}{250}$

$MR = 16 - \dfrac{x}{125}$

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At $x = 5$:

$MR = 16 - \dfrac{5}{125}$

$MR = 16 - \dfrac{1}{25}$

$MR = 16 - 0.04$

$MR = ₹\,15.96$