Q.A small manufacturer has employed 5 skilled men and 10 semi-skilled men and makes an articles in two qualities, a deluxe model and an ordinary model. The making of a deluxe requires 2 hours work by a skilled man and 2 hours work by a semi-skilled man. The ordinary model requires 1 hour by a skilled man and 3 hours by a semi-skilled man. By union rules, no man can work more than 8 hours per day. The manufacturer gains Rs 15 on the deluxe model and Rs 10 on the ordinary model. How many of each type should be made in order to maximize his total daily profit? Also, find the maximum daily 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. …