A telephone company in a town has 5000 subscribers on its list and collects fixed rent charges of Rs.3,000 per year from each subscriber. The company proposes to increase annual rent and it is believed that for every increase of one rupee in the rent, one subscriber will be discontinued. Find what increased annual rent will bring the maximum annual income to the company.
Solution
Here, the number of subscribers = 5000 and annual rental charges per subscriber = Rs.3000.
For every increase of 1 rupee in the rent, one subscriber will be discontinued.
Let the rent be increased by Rs. x.
New rental charges per year = 3000 + x
and number of subscribers after the increase in rental charges = 5000 – x. [1 M]
Let R be the annual income of the company.
Then,R =(3000+x)(5000-x)
=15000000-3000x+5000x-x2
`=15000000+2000x-x^2 and (d^2R)/dx^2=-2`
R is maximum if `"dR"/dx=0 i.e, 2000-2x=0`
`i.e if x=1000`
`((d^2R)/dx^2)_(x=1000)=-2<0`
By the second derivative test, R is maximum when x =1000.
Thus, the annual income of the company is maximum when the annual rental charges are in creased by Rs.1000.
