Q.An industry manufactures toy cars and cycles. It can invest Rs. 8900 in both of them. A car costs Rs. 450 and a cycle costs Rs. 350. It has a storage capacity of 22 items only. If its profit in Rs. 60 per car and Rs. 50 per cycle, how many of each should be manufactured so that the profit is maximum? Find also the maximum profit graphically.

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. …