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,\[R(x) = 3{x^2} + 36x + 5\]
          … (1)
Now, we know that,
⇒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.