Q.Manufacturer can sell $x$ items at a price of rupees $\left(5 - \dfrac{x}{100}\right)$ each. The cost price of $x$ items is Rs $\left(\dfrac{x}{5} + 500\right)$. Find the number of items he should sell to earn maximum profit.

Concept: Profit Maximization — Profit is revenue minus cost; maximum occurs where marginal profit is zero.

Step 1 – Write profit function

Revenue: $R(x) = x \left(5 - \frac{x}{100}\right) = 5x - \frac{x^2}{100}$

Cost: $C(x) = \frac{x}{5} + 500$

Profit: $P(x) = R(x) - C(x) = 5x - \frac{x^2}{100} - \frac{x}{5} - 500$

Step 2 – Simplify

$P(x) = \left(5 - \frac{1}{5}\right)x - \frac{x^2}{100} - 500 = \frac{24}{5}x - \frac{x^2}{100} - 500$

Step 3 – Differentiate and set to zero …

The core idea here is Profit Maximization — a classic application of derivatives in economics. Profit is simply what you earn minus what you spend. If you can write both revenue and cost as functions of the number of items $x$, then profit $P(x)$ becomes a function too. The maximum of a smooth function occurs where its derivative is zero (and the second derivative confirms it's a maximum, not a minimum).

Let’s build this step by step.

1. Write the revenue function. Revenue = (price per item) × (number of items sold). Price per item is given as $\left(5 - \frac{x}{100}\right)$. So:

2. Write the cost function. Cost of $x$ items is directly given:

3. Profit = Revenue – Cost.

Simplify:

Combine the $x$ terms: $5x - \frac{x}{5} = \frac{25x}{5} - \frac{x}{5} = \frac{24x}{5}$. So:

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

4. Differentiate $P(x)$ with respect to $x$.

5. Set $P'(x) = 0$ to find the critical point.

Multiply both sides by 50: …

Method: Optimising a Profit Function with Calculus

This method applies whenever a problem gives you a price-per-unit expression and a cost expression in terms of the number of units $x$, and asks for the quantity that earns maximum profit.

Steps

Step 1: Write Revenue and Cost as functions of $x$

Revenue is (price per unit) $\times$ (number of units sold); Cost is whatever expression the problem gives for producing/selling $x$ units. Write both explicitly:

Step 2: Form the profit function and simplify

Expand and collect like terms so $P(x)$ is a single polynomial in $x$ (typically quadratic for this kind of problem).

Step 3: Differentiate and set $P'(x) = 0$

Find $P'(x)$ term by term, then solve $P'(x) = 0$ for $x$. This locates the critical point(s) — the values of $x$ where profit is momentarily neither rising nor falling.

Step 4: Confirm it is a maximum with the second-derivative test …

Common Mistakes

Mistake 1: Maximising revenue instead of profit

Why it's wrong: The question asks for maximum profit, not maximum revenue. $R(x)$ alone ignores the cost of production, so maximising $R(x)$ gives the wrong value of $x$. Correct approach: always form $P(x) = R(x) - C(x)$ first, and differentiate the profit function, not the revenue function.

Mistake 2: Losing track of the linear cost term while forming $P(x)$

Why it's wrong: When combining $R(x) = 5x - \dfrac{x^2}{100}$ with $C(x) = \dfrac{x}{5} + 500$, it's easy to mishandle the $x$-terms from the two expressions (i.e. $5x - \dfrac{x}{5}$) and end up with the wrong linear coefficient in $P(x)$, which shifts the final critical point. Correct approach: expand $P(x) = R(x) - C(x)$ fully and combine the $x$-terms carefully ($5x - \dfrac{x}{5} = \dfrac{24x}{5}$) before differentiating.

Mistake 3: Sign or arithmetic slip differentiating $-\dfrac{x^2}{100}$ …