Q.A manufacturer produces nuts and bolts for industrial machinery. It takes 1 hour of work on machine A and 3 hours on machine B to produce a packet of nuts while it takes 3 hours on machine A and 1 hour on machine B to produce a packet of bolts. He earns a profit of Rs. 17.50 per packet on nuts and Rs. 7 per packet on bolts. How many packets of each should be produced each day so as to maximize his profit if he operates his machines for at the most 12 hours a day? Also, find the maximum profit.

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. …