A telephone company in a town has 500 subscribers on its list and collects fixed charges of Rs.300 per subscriber per year. The company proposes to increase the annual subscription and it is believed that for every increase of Rs.1 per one subscriber will discontinue the service. Find what increase will bring maximum profit?
Let us assume that company increases the annual subscription by Rs.$x$.
So, $x$ subscribes will discontinue the service.
Therefore the total revenue of company after the increment is given by
$R(x) = (500 - x)(300 + x)$
$= 15 \times {10^4} + 500x - 300x - {x^2}$
$= - {x^2} + 200x + 150000$
On differentiating both sides w.r.t. $x$, we get
${R^\prime }(x) = - 2x + 200$
Now, ${R^\prime }(x) = 0$
$\Rightarrow$ $2x = 200 \Rightarrow x = 100$
Therefore,${R^{\prime \prime }}(x) = - 2 < 0$
So, $R(x)$ is maximum when $x = 100$.
Hence, the company should increase the subscription fee by Rs.100, so that it has maximum profit.
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Application of Derivatives. Curated by Sachin Sharma. Free for all students.