Given: a telephone company in a town has 500 subscribers and collects fixed charges of Rs 300/- per subscriber per year, company increase the annual subscription and for every increase of Re 1/- one subscriber will discontinue the service
To find: the increase at which the company will get maximum profit
Explanation: company has 500 subscribers, and collects 300 per subscriber per year.
So total revenue of the company will be
R = 500×300 = 150000\-
Let the increase in annual subscription by the company be x.
Then as per the given criteria, x subscribers will discontinue the service.
Therefore the total revenue of the company after the increment is given by
R(x) = (500-x)(300+x)
⇒ R(x) = 500(300+x)-x(300+x)
⇒ R(x) = 150000+500x-300x-x2
⇒ R(x) = 150000+200x-x2
Now we will find the first derivative of the above equation, we get
R'(x) = 0+200-2x…..(i)
Now to find the critical points we will equate the first derivative to 0, i.e.,
R’(x) = 0
⇒ 200-2x = 0
⇒2x = 200
⇒ x = 100……(ii)
Now we will find the second derivative of the total revenue function, i.e., again differentiate equation (i), i.e.,
R’’(x) = 0-2<0
Hence R’’(100) is also less than 0,
Therefore, R(x) is maximum at x = 0, i.e.,
The company should the increase the subscription fee by 100 so that it will get maximum profit.
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