Q.A manufacturer produces two types of steel trunk. He has two machines, $A$ and $B$. The first type of trunk requires 3 hours on machine $A$ and 3 hours on machine $B$. The second type of trunk requires 3 hours on machine $A$ and 2 hours on machine $B$. Machines $A$ and $B$ can work at most for 18 hours and 15 hours per day respectively. He earns a profit of ₹30 and ₹25 per trunk of the first type and second type respectively. How many trunks of each type must he make each day to make the maximum profit? OR If a young man rides his motorcycle at 25 km per hour, he has to spend ₹2 per km on petrol. If he rides it at a faster speed of 40 km per hour, the petrol cost increases to ₹5 per km. He has ₹100 to spend on petrol and wishes to find the maximum distance he can travel within one hour. Express this as a linear programming problem and then solve it.

Profit Maximization

A business earns revenue from selling goods and pays out costs to make them. Profit is simply what is left over:

where $x$ is the number of units produced (or sold), $R(x)$ is the total revenue and $C(x)$ is the total cost. Profit maximization asks a very concrete question: what value of $x$ makes $P(x)$ as large as possible? In the Application of Derivatives chapter this is just a maxima-and-minima problem in disguise.

The intuition

Produce very little and you earn almost nothing; produce far too much and your costs run ahead of your revenue. Somewhere in between there is a sweet spot. As you increase production, profit first rises, then flattens, then falls — so at the top the rate of change of profit is momentarily zero. That is exactly what a derivative detects.

Finding the maximum

Treat $P(x)$ as a differentiable function and follow the standard optimisation steps:

1. Differentiate: $P'(x) = R'(x) - C'(x)$.
2. Set the derivative to zero to locate critical points: $P'(x) = 0 \ \Rightarrow\ R'(x) = C'(x)$.
3. Confirm it is a maximum with the second-derivative test: profit is maximised at a critical point $x^{*}$ where

A quick illustration

Suppose $P(x) = 100x - 2x^2$ (in rupees). Then $P'(x) = 100 - 4x$. Setting $P'(x) = 0$ gives $x = 25$. Since $P''(x) = -4 < 0$, the profit is indeed greatest at $x = 25$ units. …