मराठी

An online delivery company in a city has 5,000 subscribers and collects annual subscription fees of ₹ 300 per subscriber for unlimited free deliveries. The company wishes to increase the annual - Mathematics

Advertisements
Advertisements

प्रश्न

An online delivery company in a city has 5,000 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.

Based on the given information, answer the following questions:

  1. How many subscribers will discontinue after an increase of ₹ x in annual fee?
  2. If R(x) denotes the total revenue collected after the increase of ₹ x in subscription fee, express R(x) as a function of x.
    1. Find the value of x for which R(x) is maximum.
      OR
    2. Find the sub-intervals of (0, 5000) in which R(x) is increasing and decreasing.
घटनेचा अभ्यास
Advertisements

उत्तर

(i)

For every ₹ 1 increase, 10 subscribers discontinue.

For ₹ x increase:

10x subscribers will discontinue.

(ii)

New subscription fee: 300 + x

Remaining subscribers: 5000 − 10x

So revenue will be,

R(x) = (300 + x)(5000 − 10x)

= 300(5000 − 10x) + x(5000 − 10x)

= 1500000 − 3000x + 5000x − 10x2

= 1500000 + 2000x − 10x2

∴ R(x) = −10x2 + 2000x + 1500000

(iii) (a)

R(x) = −10x2 + 2000x + 1500000

⇒ Differentiate:

R′(x) = −20x + 2000

⇒ For maximum:

R′(x) = 0

−20x + 2000 = 0

∴ x = 100

So, revenue is at its maximum when the annual fee is increased by ₹ 100.

OR

(iii) (b)

R′(x) = −20x + 2000

⇒ For increasing:

R′(x) > 0

−20x + 2000 > 0

x < 100

⇒ For decreasing:

R′(x) < 0

x > 100

⇒ Hence, in (0, 5000):

R(x) is increasing on (0, 100) and R(x) is decreasing on (100, 5000).

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2025-2026 (March) 65/1/1
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×