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A telephone company in a town has 500 subscribers on its list and collects fixed charges of $₹ 300$ per subscriber per year. The company proposes to increase the annual subscription and it is believed that for every increase of ₹ 1 per one subscriber will discontinue the service. Find what increase will bring maximum profit?
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Verified Answer
Consider that company increases the annual subscription by $₹ x$.
So, $\mathrm{x}$ subscribes will discontinue the service.
$\therefore$ Total revenue of company after the increment is given by
$$
\begin{aligned}
&R(x)=(500-x)(300+x) \\
&=-x^2+200 x+150000
\end{aligned}
$$
On differentiating both sides w.r.t. $x$, we get
$$
R^{\prime}(x)=-2 x+200
$$
For, $R^{\prime}(x)=0,2 x=200 \Rightarrow x=100$
also $R^{\prime \prime}(\mathrm{x})=-2 < 0$
So, $\mathrm{R}(\mathrm{x})$ is maximum when $\mathrm{x}=100$
Hence the company should increase the subscription fee by $₹ 100$, so that it has maximum profit.
$\Rightarrow \mathrm{x}=100$ also $\mathrm{R}^{\prime \prime}(\mathrm{x})=-2 < 0$
So, $\mathrm{R}(\mathrm{x})$ is maximum when $x=100$
Hence the company should increase the subscription fee by $₹ 100$, so that it has maximum profit.
So, $\mathrm{x}$ subscribes will discontinue the service.
$\therefore$ Total revenue of company after the increment is given by
$$
\begin{aligned}
&R(x)=(500-x)(300+x) \\
&=-x^2+200 x+150000
\end{aligned}
$$
On differentiating both sides w.r.t. $x$, we get
$$
R^{\prime}(x)=-2 x+200
$$
For, $R^{\prime}(x)=0,2 x=200 \Rightarrow x=100$
also $R^{\prime \prime}(\mathrm{x})=-2 < 0$
So, $\mathrm{R}(\mathrm{x})$ is maximum when $\mathrm{x}=100$
Hence the company should increase the subscription fee by $₹ 100$, so that it has maximum profit.
$\Rightarrow \mathrm{x}=100$ also $\mathrm{R}^{\prime \prime}(\mathrm{x})=-2 < 0$
So, $\mathrm{R}(\mathrm{x})$ is maximum when $x=100$
Hence the company should increase the subscription fee by $₹ 100$, so that it has maximum profit.
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