Question

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

Answer 1

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,$$R(x)=3x^2+36x+5$$
          … (1)
Now, we know that,
⇒Marginal revenue $={\displaystyle \frac{dR(x)}{dx}}$
Therefore, differentiating equation (1) with respect to $x$ , we get,
$${\displaystyle \frac{dR(x)}{dx}}=3{\displaystyle \frac{d{x}^2}{dx}}+36{\displaystyle \frac{dx}{dx}}+5{\displaystyle \frac{d1}{dx}}$$
$$\Rightarrow {\displaystyle \frac{dR(x)}{dx}}=3(2x)+36x+0$$
$$\Rightarrow {\displaystyle \frac{dR(x)}{dx}}=6x+36$$
It is given in the question that we have to calculate the marginal revenue at $$x=15$$
Therefore, Marginal revenue $$=6(15)+36$$
$$\therefore {\displaystyle \frac{dR(x)}{dx}}=126$$
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.