The question
An online delivery company in a city has 5000 subscribers and collects annual subscription fees of ₹ 300 per subscriber for unlimited free deliveries.
The company wishes to increase the annual subscription fee. It is predicted that, for every increase of ₹ 1, ten subscribers will discontinue. Assume that the company increased the annual fee by ₹ x x x
Based on the given information, answer the following questions :
(i) How many subscribers will discontinue after an increase of ₹ x x x (1 mark)
(ii) If R ( x ) R(x) R ( x ) x x x R ( x ) R(x) R ( x ) x x x (1 mark)
(iii) (a) Find the value of x x x R ( x ) R(x) R ( x ) (2 marks)
OR
(iii) (b) Find the sub-intervals of ( 0 , 5000 ) (0, 5000) ( 0 , 5000 ) R ( x ) R(x) R ( x ) (2 marks)
Exam answer Show all explanations
(i) For every increase of ₹ 1, 10 subscribers discontinue.
Number of subscribers who discontinue = 10 x = 10x = 10 x
Why?
The loss is 10 subscribers for each rupee of increase, so an increase of x x x 10 × x 10 \times x 10 × x
(ii) New fee = 300 + x = 300 + x = 300 + x = 5000 − 10 x = 5000 - 10x = 5000 − 10 x
Why?
The fee goes up from ₹ 300 by x x x 10 x 10x 10 x
R ( x ) = ( 5000 − 10 x ) ( 300 + x ) R(x) = (5000 - 10x)(300 + x) R ( x ) = ( 5000 − 10 x ) ( 300 + x )
Why?
Revenue is the number of paying subscribers times the fee each one pays.
(iii) (a)
R ( x ) = 1500000 + 5000 x − 3000 x − 10 x 2 \begin{aligned} R(x) &= 1500000 + 5000x \\ &\quad - 3000x - 10x^2 \end{aligned} R ( x ) = 1500000 + 5000 x − 3000 x − 10 x 2 R ( x ) = 1500000 + 2000 x − 10 x 2 R ′ ( x ) = 2000 − 20 x \begin{aligned} R(x) &= 1500000 + 2000x - 10x^2 \\ R'(x) &= 2000 - 20x \end{aligned} R ( x ) R ′ ( x ) = 1500000 + 2000 x − 10 x 2 = 2000 − 20 x Why?
Expand the product term by term:
5000 × 300 = 1500000 5000 × x = 5000 x − 10 x × 300 = − 3000 x − 10 x × x = − 10 x 2 \begin{aligned} 5000 \times 300 &= 1500000 \\ 5000 \times x &= 5000x \\ -10x \times 300 &= -3000x \\ -10x \times x &= -10x^2 \end{aligned} 5000 × 300 5000 × x − 10 x × 300 − 10 x × x = 1500000 = 5000 x = − 3000 x = − 10 x 2 Collect the x x x 5000 x − 3000 x = 2000 x 5000x - 3000x = 2000x 5000 x − 3000 x = 2000 x
Then differentiate each term: the constant gives 0, 2000 x 2000x 2000 x − 10 x 2 -10x^2 − 10 x 2 − 20 x -20x − 20 x
d d x ( 1500000 ) = 0 d d x ( 2000 x ) = 2000 d d x ( − 10 x 2 ) = − 10 ( 2 x ) = − 20 x \begin{aligned} \frac{d}{dx}(1500000) &= 0 \\ \frac{d}{dx}(2000x) &= 2000 \\ \frac{d}{dx}(-10x^2) &= -10(2x) = -20x \end{aligned} d x d ( 1500000 ) d x d ( 2000 x ) d x d ( − 10 x 2 ) = 0 = 2000 = − 10 ( 2 x ) = − 20 x R ′ ( x ) = 0 ⇒ x = 100 R'(x) = 0 \Rightarrow x = 100 R ′ ( x ) = 0 ⇒ x = 100 Why?
At a maximum the tangent is horizontal, so we look where R ′ ( x ) = 0 R'(x) = 0 R ′ ( x ) = 0 20 x = 2000 20x = 2000 20 x = 2000 x = 100 x = 100 x = 100
2000 − 20 x = 0 2000 = 20 x x = 2000 20 = 100 \begin{aligned} 2000 - 20x &= 0 \\ 2000 &= 20x \\ x &= \frac{2000}{20} = 100 \end{aligned} 2000 − 20 x 2000 x = 0 = 20 x = 20 2000 = 100 Watch out: R ′ ( x ) = 0 R'(x) = 0 R ′ ( x ) = 0
R ′ ′ ( x ) = − 20 < 0 at x = 100 R''(x) = -20 < 0 \text{ at } x = 100 R ′′ ( x ) = − 20 < 0 at x = 100 Why?
Second derivative test: if R ′ ( c ) = 0 R'(c) = 0 R ′ ( c ) = 0 R ′ ′ ( c ) < 0 R''(c) < 0 R ′′ ( c ) < 0 R R R c c c
Differentiate R ′ ( x ) = 2000 − 20 x R'(x) = 2000 - 20x R ′ ( x ) = 2000 − 20 x
R ′ ′ ( x ) = d d x ( 2000 ) − d d x ( 20 x ) = 0 − 20 = − 20 \begin{aligned} R''(x) &= \frac{d}{dx}(2000) - \frac{d}{dx}(20x) \\ &= 0 - 20 = -20 \end{aligned} R ′′ ( x ) = d x d ( 2000 ) − d x d ( 20 x ) = 0 − 20 = − 20 Here R ′ ′ ( x ) = − 20 R''(x) = -20 R ′′ ( x ) = − 20 x x x x = 100 x = 100 x = 100
∴ R ( x ) \therefore\ R(x) ∴ R ( x ) x = 100 x = 100 x = 100
(iii) (b)
R ′ ( x ) = 2000 − 20 x = 20 ( 100 − x ) R'(x) = 2000 - 20x = 20(100 - x) R ′ ( x ) = 2000 − 20 x = 20 ( 100 − x ) R ′ ( x ) = 0 ⇒ x = 100 R'(x) = 0 \Rightarrow x = 100 R ′ ( x ) = 0 ⇒ x = 100 Why?
The sign of R ′ ( x ) R'(x) R ′ ( x ) R R R R ′ ( x ) R'(x) R ′ ( x ) 20 ( 100 − x ) 20(100 - x) 20 ( 100 − x )
Take out the common factor 20: 2000 = 20 × 100 2000 = 20 \times 100 2000 = 20 × 100 20 x = 20 × x 20x = 20 \times x 20 x = 20 × x 20 ( 100 − x ) = 0 20(100 - x) = 0 20 ( 100 − x ) = 0 100 − x = 0 100 - x = 0 100 − x = 0 x = 100 x = 100 x = 100
The point x = 100 x = 100 x = 100 R ′ ( x ) = 0 R'(x) = 0 R ′ ( x ) = 0 ( 0 , 5000 ) (0, 5000) ( 0 , 5000 ) ( 0 , 100 ) (0, 100) ( 0 , 100 ) ( 100 , 5000 ) (100, 5000) ( 100 , 5000 )
In ( 0 , 100 ) (0, 100) ( 0 , 100 ) R ′ ( x ) > 0 R'(x) > 0 R ′ ( x ) > 0 R ( x ) R(x) R ( x ) ( 0 , 100 ) (0, 100) ( 0 , 100 )
Why?
For x < 100 x < 100 x < 100 100 − x > 0 100 - x > 0 100 − x > 0 x = 50 x = 50 x = 50 R ′ ( 50 ) = 20 ( 50 ) = 1000 > 0 R'(50) = 20(50) = 1000 > 0 R ′ ( 50 ) = 20 ( 50 ) = 1000 > 0
R ′ ( 50 ) = 20 ( 100 − 50 ) = 20 ( 50 ) = 1000 \begin{aligned} R'(50) &= 20(100 - 50) \\ &= 20(50) = 1000 \end{aligned} R ′ ( 50 ) = 20 ( 100 − 50 ) = 20 ( 50 ) = 1000 A function with a positive derivative on an interval is increasing there.
In ( 100 , 5000 ) (100, 5000) ( 100 , 5000 ) R ′ ( x ) < 0 R'(x) < 0 R ′ ( x ) < 0 R ( x ) R(x) R ( x ) ( 100 , 5000 ) (100, 5000) ( 100 , 5000 )
Why?
For x > 100 x > 100 x > 100 100 − x < 0 100 - x < 0 100 − x < 0 R ′ ( 200 ) = 20 ( − 100 ) R'(200) = 20(-100) R ′ ( 200 ) = 20 ( − 100 ) − 2000 < 0 -2000 < 0 − 2000 < 0
R ′ ( 200 ) = 20 ( 100 − 200 ) = 20 ( − 100 ) = − 2000 \begin{aligned} R'(200) &= 20(100 - 200) \\ &= 20(-100) = -2000 \end{aligned} R ′ ( 200 ) = 20 ( 100 − 200 ) = 20 ( − 100 ) = − 2000 ∴ R ( x ) \therefore\ R(x) ∴ R ( x ) ( 0 , 100 ) (0, 100) ( 0 , 100 ) ( 100 , 5000 ) (100, 5000) ( 100 , 5000 )
Where this comes from Chapter 6, Application of Derivatives : Revenue: maximum, increasing and decreasing.
Question type: (iii)(a) Word problems ; (iii)(b) Find where a function increases and decreases
How to spot it: click to see (iii)(a) Type 10. Word problems: the biggest box, the least cost
Spot it: a real object and the words "maximum", "minimum", "largest" or "least".
Here: The revenue R ( x ) = ( 5000 − 10 x ) ( 300 + x ) R(x) = (5000 - 10x)(300 + x) R ( x ) = ( 5000 − 10 x ) ( 300 + x ) R ′ ( x ) = 2000 − 20 x = 0 R'(x) = 2000 - 20x = 0 R ′ ( x ) = 2000 − 20 x = 0 x = 100 x = 100 x = 100 R ′ ′ = − 20 R'' = -20 R ′′ = − 20
(iii)(b) Type 5. Find where a function increases and decreases
Spot it: "Find the intervals in which f f f
Here: Write R ′ ( x ) = 20 ( 100 − x ) R'(x) = 20(100 - x) R ′ ( x ) = 20 ( 100 − x ) x = 100 x = 100 x = 100 ( 0 , 5000 ) (0, 5000) ( 0 , 5000 ) ( 0 , 100 ) (0, 100) ( 0 , 100 ) R ′ R' R ′ ( 100 , 5000 ) (100, 5000) ( 100 , 5000 )
Quick lesson · Worked NCERT solutions · Interactive lesson