Q38
2 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).
first derivative testsecond derivative testmaxima of a functionincreasing and decreasing functionscritical pointR'(x)=0

Marking Scheme

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

Analysis & 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.

Common Mistakes

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

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.

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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.