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  • 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 Re 1/- one subscriber will

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 Re 1/- one subscriber will discontinue the service. Find what increase will bring maximum profit?

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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 best increase amount for the company to earn maximum profit

Explanation: company has 500 subscribers, and collects 300 per subscriber per year.

Let x as the increase in annual subscription by the company

As per the question, the number of subscribers to discontinue the service will be x

The total revenue earned after the increment would be calculated by,

\\{R(x) = (500-x)(300+x)}\\ { \Rightarrow R(x) = 500(300+x)-x(300+x)}\\ { \Rightarrow R(x) = 150000+500x-300x-x\textsuperscript{2}}\\ { \Rightarrow R(x) = 150000+200x-x\textsuperscript{2}}\\

We need to calculate the first derivative of the above equation,

\\ \mathrm{R}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(150000+200 \mathrm{x}-\mathrm{x}^{2}\right)}{\mathrm{d} \mathrm{x}} \\ \mathrm{R}^{\prime}(\mathrm{x})=\frac{\mathrm{d}(150000)}{\mathrm{dx}}+\frac{\mathrm{d}(200 \mathrm{x})}{\mathrm{d} \mathrm{x}}-\frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{d} \mathrm{x}}

R'(x) = 0+200-2x \ldots ..(i)

The critical point is calculated by equating the first derivative with 0,

\\{R'(x) = 0}\\ { \Rightarrow 200-2x = 0}\\ { \Rightarrow 2x = 200}\\ { \Rightarrow x = 100 \ldots \ldots (ii)}\\

Then we calculate the second derivative of the total revenue function, i.e., again differentiate equation (i), i.e.,

\\ \mathrm{R}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}(200-2 \mathrm{x})}{\mathrm{dx}} \\ \mathrm{R}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}(200)}{\mathrm{dx}}-\frac{\mathrm{d}(2 \mathrm{x})}{\mathrm{d} \mathrm{x}} \\ \mathrm{R}^{\prime \prime}(\mathrm{x})=0-2<0

Hence R’’(100) is also less than 0,

Therefore, R(x) is maximum at x = 0, i.e.,

Thus, the required increase on the subscription fee for the company to make profit is by Rs 100.

Posted by

infoexpert22

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