Q.An online delivery company has 5000 customers in a city and charges each customer ₹ 300 per annum for unlimited free deliveries. The company wants to increase its annual subscription fee. It is estimated that for every ₹ 1 increase, 10 members will leave. Let the company increase the annual fee by ₹ $x$.

The key idea is Profit Maximization — the company’s revenue is the product of price per customer and the number of customers, and we find the price increase that maximizes this revenue.

Step 1: Express the number of customers after an increase of ₹ $x$.

Initial customers = 5000. For each ₹1 increase, 10 leave. So customers left = $5000 - 10x$.

Step 2: Express the new annual fee per customer.

Original fee = ₹300. After increase by ₹$x$, fee = $300 + x$.

Step 3: Write the revenue function $R(x)$.

Expand: …

Why profit maximisation works this way

When a business changes its price, two things happen in opposite directions. A higher price means more revenue per customer, but it also drives some customers away. The trick is to find the sweet spot where the gain from the higher price is just balanced by the loss from fewer customers.

Here, the company starts with 5000 customers at ₹ 300 each. For every ₹ 1 increase, 10 customers leave. So if the fee goes up by ₹ $x$, the new fee is ₹ $(300 + x)$ and the number of customers drops to $(5000 - 10x)$.

The total annual revenue (which is also the profit here, since we aren't given any costs) is:

This is a quadratic expression. When you expand it, the $x^2$ term will have a negative coefficient, so the graph is a downward-opening parabola. The maximum profit occurs at the vertex.

Step-by-step solution

1. Write the profit function

Let $P(x)$ be the annual profit when the fee is increased by ₹ $x$.

2. Expand to standard quadratic form

So $P(x) = -10x^2 + 2000x + 15,00,000$.

3. Find the vertex

Here $a = -10$, $b = 2000$, $c = 15,00,000$.

So the optimal increase is ₹ 100. …