Q.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.

Concept: Profit Maximization — profit is revenue minus cost; here cost is fixed, so we maximise revenue.

Let $x$ be the number of Re 1 increases.

New subscription charge = $300 + x$ rupees.

Number of subscribers = $500 - x$.

Revenue $R(x) = (300 + x)(500 - x) = 150000 + 200x - x^2$. …

Why this approach works

Profit is simply total revenue minus fixed costs. Here, the company’s only cost is negligible (or zero for the purpose of maximisation), so maximising profit is equivalent to maximising total revenue. The twist is that raising the price drives away subscribers — a trade-off every business faces. We need to find the sweet spot where the gain from a higher price is exactly balanced by the loss from fewer customers.

Let the current situation be:

• Price per subscriber = Rs 300
• Number of subscribers = 500
• Current revenue = $300 \times 500 = \text{Rs } 1,50,000$

If the company increases the subscription by Rs $x$ (where $x \geq 0$), then:

• New price per subscriber = $300 + x$
• For every Re 1 increase, one subscriber leaves. So number of subscribers lost = $x$
• New number of subscribers = $500 - x$

Revenue $R(x)$ becomes:

We want the $x$ that makes $R(x)$ as large as possible.

Step-by-step solution

1. Write the revenue function

Expand:

This is a quadratic in $x$, with a negative coefficient for $x^2$. That means its graph is an upside-down parabola — it has a single maximum point.

2. Find the vertex of the parabola

For any quadratic $ax^2 + bx + c$, the maximum (or minimum) occurs at $x = -\frac{b}{2a}$. Here:

So:

Thus, the revenue is maximised when the increase is Rs 100. …

Method: Optimizing Revenue Under a Linear Price-Demand Trade-off

This is the standard technique for problems where, for every unit increase in price, a fixed number of units or customers is lost — a classic linear trade-off that turns into a quadratic revenue function.

Steps

Step 1: Define the increment variable and write the new price and new quantity

Let $x$ be the number of price increments applied. Write the new price and the new quantity as linear functions of $x$: price goes up by $x$ times the increment, quantity goes down by $x$ times the loss-per-increment.

Step 2: Form the revenue function and expand

Multiply out — this always collapses to a quadratic in $x$ with a negative leading coefficient, since price and quantity move in opposite directions, guaranteeing a single maximum.

Step 3: Maximize the quadratic …

Common Mistakes

Mistake 1: Assuming the maximum-revenue increase is Rs 250, "by symmetry" of the 500 subscribers

Why it's wrong: that reasoning would apply if revenue were simply $x(500-x)$, but the actual revenue function is $(300+x)(500-x)$ — the extra $300$ shifts the vertex away from the naive midpoint. Correct approach: always expand the full product into standard quadratic form ($150000+200x-x^2$) and apply $x=-b/2a$ to that quadratic, rather than guessing from the subscriber count alone.

Mistake 2: A sign error while expanding $(300+x)(500-x)$ …