Question 371 markSection E

If R(x)R(x) denotes the total revenue collected after the increase of Rs x in subscription fee, express R(x)R(x) as a function of x.

Application of Derivatives
Application of Derivatives — Setting up a Revenue Function

Official answer

Revenue = (number of remaining subscribers)×(new fee per subscriber)(\text{number of remaining subscribers}) \times (\text{new fee per subscriber}).


R(x)=(5000−10x)(300+x)R(x) = (5000 - 10x)(300 + x)

Answer keywords:revenue functionproduct of price and quantityalgebraic modelling5000-10x300+x

Marking scheme

  1. 1.1 mark: correct function R(x)=(5000−10x)(300+x)R(x) = (5000-10x)(300+x), or its expanded form −10x2+2000x+1500000-10x^2+2000x+1500000.

Hint

Revenue = (5000−10x)(5000 - 10x) subscribers remaining × (300+x)(300 + x) new fee per subscriber.

Quick oral answer

R(x)R(x) is the product of the remaining subscribers (5000−10x)(5000-10x) and the new fee (300+x)(300+x), i.e., R(x)=(5000−10x)(300+x)R(x) = (5000-10x)(300+x).

Common mistakes

  1. 1.Using the original 5000 subscribers instead of the reduced (5000−10x)(5000-10x) in the revenue product.
  2. 2.Writing the new fee as just 'x' instead of '300+x300+x' (the increase, not the total new fee).

Analysis and explanation

This step converts the two verbal quantities — remaining subscribers and new fee — into a single revenue function, which becomes the object of optimisation in the next parts.


Concept

  • Remaining subscribers after the increase = 5000−10x5000 - 10x (original count minus those who discontinue).
  • New fee per subscriber = 300+x300 + x.
  • Total revenue = (remaining subscribers) × (new fee) = (5000−10x)(300+x)(5000-10x)(300+x).

Exam trap

  • Students sometimes multiply the original 5000 subscribers by the new fee, forgetting to subtract the 10x10x who leave — this is the single most common error in this case study.

Interesting facts

This is a textbook example of the classic 'demand-revenue' optimisation problem, structurally identical to problems economists use to find revenue-maximising prices for real subscription services.

Expanding R(x)R(x) gives a downward parabola −10x2+2000x+1500000-10x^2+2000x+1500000, whose vertex directly gives the revenue-maximising price increase — a link between algebra (quadratics) and calculus (derivatives) tested later in the case study.

Question details

Q37 of 44 questions

Board
CBSE
Class
Class 12
Subject
Mathematics
Year
2026
Section
Section E
Marks
1
Chapter
Application of Derivatives
Topic
Application of Derivatives — Setting up a Revenue Function
Bloom's level
apply
Syllabus topic
Unit 6: Application of Derivatives — Maxima and Minima (Case Study: Modelling Revenue as a Function of Price Increase)
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why is R(x)R(x) a product of two linear expressions?

Because total revenue is always (price per unit)×(number of units sold)(\text{price per unit}) \times (\text{number of units sold}), and here both the price and the number of subscribers change linearly with x.