Question

# The total revenue in Rupees received from the sale of $x$ units of a product is given by $R(x) = 3{x^2} + 36x + 5$. The marginal revenue, when $x = 15$ is,(a) $116$(b) $96$(c) $90$(d) $126$

Hint: Differentiate the given equation carefully without missing any term in between. ALSO Marginal revenue is the derivative of total revenue with respect to demand.

We have the given equation as,$R(x) = 3{x^2} + 36x + 5$
â€¦ (1)
Now, we know that,
â‡’Marginal revenue $= \dfrac{{dR(x)}}{{dx}}$
Therefore, differentiating equation (1) with respect to $x$ , we get,
$\dfrac{{dR(x)}}{{dx}} = 3\dfrac{{d{x^2}}}{{dx}} + 36\dfrac{{dx}}{{dx}} + 5\dfrac{{d1}}{{dx}}$
$\Rightarrow \dfrac{{dR(x)}}{{dx}} = 3(2x) + 36x + 0$
$\Rightarrow \dfrac{{dR(x)}}{{dx}} = 6x + 36$
It is given in the question that we have to calculate the marginal revenue at $x = 15$
Therefore, Marginal revenue $= 6(15) + 36$
$\therefore \dfrac{{dR(x)}}{{dx}} = 126$
Hence, the marginal revenue at $x = 15$ is $126$.
So, the required solution is (d) $126$.

Note: To solve these types of problems, simply differentiate the given equation and substitute the value of the given variable to obtain an optimum solution.