Question 382 marksSection E

(a) Find the value of x for which R(x)R(x) is maximum. OR (b) Find the sub-intervals of (0,5000)(0, 5000) in which R(x)R(x) is increasing and decreasing.

Application of Derivatives
Application of Derivatives — Maxima/Minima and Increasing-Decreasing Functions

Official answer

Using R(x)=(5000−10x)(300+x)=−10x2+2000x+1500000R(x) = (5000-10x)(300+x) = -10x^2 + 2000x + 1500000 from the previous part.


Part (a): Maximum revenue

  • R′(x)=−20x+2000R'(x) = -20x + 2000; setting R′(x)=0R'(x) = 0 gives x=100x = 100.
  • R′′(x)=−20<0R''(x) = -20 < 0, confirming a maximum. So R(x)R(x) is maximum at x=100x = 100.

Part (b): Increasing/decreasing intervals

  • R′(x)=−20x+2000>0R'(x) = -20x + 2000 > 0 for x<100x < 100, so R(x)R(x) is increasing on (0,100)(0, 100).
  • R′(x)<0R'(x) < 0 for x>100x > 100, so R(x)R(x) is decreasing on (100,5000)(100, 5000).
Answer keywords:first derivative testsecond derivative testmaxima of a functionincreasing and decreasing functionscritical pointR'(x)=0

Marking scheme

  1. 1.Part (a) — 1 mark: correct R′(x)=−20x+2000R'(x) = -20x+2000 and setting it to 0 to get x=100x=100.
  2. 2.Part (a) — 1 mark: verifying via R′′(x)=−20<0R''(x)=-20<0 (or sign change) that x=100x=100 gives a maximum.
  3. 3.Part (b) — 1 mark: correctly identifying R(x)R(x) is increasing on (0,100)(0,100).
  4. 4.Part (b) — 1 mark: correctly identifying R(x)R(x) is decreasing on (100,5000)(100,5000).

Hint

Differentiate R(x)=−10x2+2000x+1500000R(x)=-10x^2+2000x+1500000, set R′(x)=0R'(x)=0 for the maximum, and study the sign of R′(x)R'(x) on either side of that point for increasing/decreasing behaviour.

Quick oral answer

R′(x)=−20x+2000=0R'(x)=-20x+2000=0 gives x=100x=100 as the revenue-maximising increase, with R(x)R(x) increasing on (0,100)(0,100) and decreasing on (100,5000)(100,5000).

Common mistakes

  1. 1.Finding the critical point x=100x=100 but not confirming it's a maximum using the second derivative test.
  2. 2.Writing incorrect or overlapping intervals for increasing/decreasing behaviour around x=100x=100.
  3. 3.Arithmetic slip while expanding (5000−10x)(300+x)(5000-10x)(300+x), leading to a wrong R′(x)R'(x).

Analysis and explanation

This part directly applies the first and second derivative tests to the revenue function built in the previous two parts.


Concept

  • R(x)=−10x2+2000x+1500000R(x) = -10x^2+2000x+1500000 is a downward parabola; its critical point (found from R′(x)=0R'(x)=0) is where the maximum occurs.
  • The sign of R′(x)R'(x) on either side of the critical point tells us where R(x)R(x) increases or decreases — positive derivative means increasing, negative means decreasing.

Exam trap

  • Students sometimes forget to verify the nature of the critical point using R′′(x)R''(x) (second derivative test) or a sign chart of R′(x)R'(x), simply assuming any critical point is automatically a maximum.
  • In part (b), the open interval (0,5000)(0,5000) must be split precisely at x=100x=100 — writing overlapping or incorrect intervals (e.g., including x=100x=100 in both) loses marks.

Real-world relevance

  • This is precisely how real companies (subscription services, ride-hailing apps, airlines) determine optimal price increases using calculus-based revenue optimisation — a direct application of the derivative test taught here.

Interesting facts

Because R(x)R(x) is a downward parabola, its vertex (revenue-maximising point) can also be found purely algebraically as x=−b2a=−20002×−10=100x = -\frac{b}{2a} = \frac{-2000}{2\times-10} = 100, matching the calculus answer exactly — a nice cross-check between algebra and calculus.

This exact optimal-pricing technique (maximise revenue = price×quantity\text{price} \times \text{quantity}, where quantity is a linear function of price) is the foundation of 'yield management' used by airlines and subscription services worldwide.

Question details

Q38 of 44 questions

Board
CBSE
Class
Class 12
Subject
Mathematics
Year
2026
Section
Section E
Marks
2
Chapter
Application of Derivatives
Topic
Application of Derivatives — Maxima/Minima and Increasing-Decreasing Functions
Bloom's level
apply
Syllabus topic
Unit 6: Application of Derivatives — First Derivative Test for Maxima, Increasing and Decreasing Functions
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

How do we know x=100x=100 gives a maximum and not a minimum?

Since R′′(x)=−20R''(x) = -20, which is negative, the critical point x=100x=100 corresponds to a maximum by the second derivative test.

Why is the interval split exactly at x=100x=100?

Because R′(x)R'(x) changes sign exactly at x=100x=100 (from positive to negative), which is the boundary between increasing and decreasing behaviour.