Question:

The total revenue in Rupees received from the sale of \(x\) units of a product is given by \(R(x)=13x^2+26x+15\).The marginal revenue, when \(x=7\) is

Updated On: Dec 26, 2024
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Solution and Explanation

The correct answer is Rs 208.
Marginal revenue is the rate of change of total revenue with respect to the number of units sold. 
\(∴Marginal Revenue (MR)=\frac{dR}{dx}=13(2x)+26=26x+26\)
When \(x = 7, MR = 26(7) + 26 = 182 + 26 = 208\)
Hence, the required marginal revenue is Rs 208.
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Concepts Used:

Application of Derivatives

Various Applications of Derivatives-

Rate of Change of Quantities:

If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by 

\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)

This is also known to be as the Average Rate of Change.

Increasing and Decreasing Function:

Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

  • If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
  • Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
  • The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.

Read More: Application of Derivatives