Question

A private telephone company serving a small community makes a profit of Rs. 12.00 per subscriber, if it has 725 subscribers. It decides to reduce the rate by a fixed sum for each subscriber over 725, thereby reducing the profit by 1 paise per subscriber. Thus, there will be profit of Rs. 11.99 on each of the 726 subscribers, Rs. 11.98 on each of the 727 subscribers, etc. What is the number of subscribers which will give the company the maximum profit?

Answer 1

962 or 963

Answer 2

The problem asks us to find the number of subscribers that will maximize the company's profit. This is an optimization problem that can be solved using differential calculus.

1. Define Variables and Formulate the Profit Function:

Let the initial number of subscribers be $N_0 = 725$.

Let the initial profit per subscriber be $P_0 = Rs. 12.00$.

The company reduces the rate by 1 paise (Rs. 0.01) for each subscriber over 725.

Let $x$ be the number of subscribers over 725.

So, the total number of subscribers $N$ will be:

$N = N_0 + x = 725 + x$

For $x$ additional subscribers, the profit per subscriber $P_s$ is reduced by $x$ times 1 paise.

So, the profit per subscriber $P_s$ becomes:

$P_s = P_0 - 0.01x = 12 - 0.01x$

The total profit $P(x)$ is the product of the total number of subscribers and the profit per subscriber:

$P(x) = N \times P_s$

$P(x) = (725 + x)(12 - 0.01x)$

2. Expand and Simplify the Profit Function:

$P(x) = 725 \times 12 + 725 \times (-0.01x) + x \times 12 + x \times (-0.01x)$

$P(x) = 8700 - 7.25x + 12x - 0.01x^2$

$P(x) = -0.01x^2 + (12 - 7.25)x + 8700$

$P(x) = -0.01x^2 + 4.75x + 8700$

3. Differentiate the Profit Function:

To find the maximum profit, we need to find the critical points by taking the first derivative of $P(x)$ with respect to $x$ and setting it to zero:

$\frac{dP}{dx} = \frac{d}{dx}(-0.01x^2 + 4.75x + 8700)$

$\frac{dP}{dx} = -0.01(2x) + 4.75 + 0$

$\frac{dP}{dx} = -0.02x + 4.75$

4. Find the Value of x for Maximum Profit:

Set the derivative to zero to find the value of $x$ that maximizes profit:

$-0.02x + 4.75 = 0$

$0.02x = 4.75$

$x = \frac{4.75}{0.02}$

$x = \frac{475}{2}$

$x = 237.5$

5. Consider Integer Number of Subscribers:

Since $x$ represents the number of additional subscribers, it must be an integer. The profit function $P(x) = -0.01x^2 + 4.75x + 8700$ is a downward-opening parabola, meaning its maximum occurs at the vertex $x = 237.5$. For integer values of $x$, the maximum profit will occur at the integers closest to $237.5$, which are $x = 237$ and $x = 238$.

Let's calculate the total profit for both these values of $x$:

• Case 1: x = 237

  Number of subscribers $N = 725 + 237 = 962$

  Profit per subscriber $P_s = 12 - 0.01 \times 237 = 12 - 2.37 = Rs. 9.63$

  Total Profit $P(237) = 962 \times 9.63 = Rs. 9262.86$

• Case 2: x = 238

  Number of subscribers $N = 725 + 238 = 963$

  Profit per subscriber $P_s = 12 - 0.01 \times 238 = 12 - 2.38 = Rs. 9.62$

  Total Profit $P(238) = 963 \times 9.62 = Rs. 9262.86$

Both $x = 237$ and $x = 238$ yield the same maximum profit of Rs. 9262.86.

Therefore, the company will achieve maximum profit with either 962 or 963 subscribers.

Explanation of the solution:

1. Define $x$ as the number of subscribers over 725.
2. Formulate the total number of subscribers as $(725 + x)$.
3. Formulate the profit per subscriber as $(12 - 0.01x)$.
4. Write the total profit function $P(x) = (725 + x)(12 - 0.01x)$.
5. Expand $P(x)$ to $P(x) = -0.01x^2 + 4.75x + 8700$.
6. Differentiate $P(x)$ with respect to $x$: $dP/dx = -0.02x + 4.75$.
7. Set $dP/dx = 0$ to find the critical point: $x = 237.5$.
8. Since $x$ must be an integer, evaluate the profit for $x = 237$ and $x = 238$.
9. Both $x = 237$ and $x = 238$ result in the same maximum profit.
10. Calculate the total number of subscribers for each $x$: $725 + 237 = 962$ and $725 + 238 = 963$.