Calculus in Context – Finding Maximum Profit

Calculus is often seen as an abstract mathematical tool, but in reality, it’s one of the most powerful methods for solving real business and economic problems. One of the clearest examples is finding maximum profit — where differentiation turns raw data into decision-making power.

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The Concept

Profit depends on revenue and cost:

Profit=Revenue−Cost

If revenue and cost each depend on the number of units sold ($x$), then profit is a function of $x$. The goal is to find the value of $x$ that gives the greatest profit.

By differentiating the profit function with respect to $x$, we can find where the slope = 0, meaning profit stops increasing — the maximum point.

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Example

Suppose a company’s profit function is:

$$P(x)=-2x^2+40x-100$$

Differentiate to find the turning point:

$$\frac{dP}{dx}=-4x+40$$

Set this to zero to find the maximum:

$$-4x+40=0\Rightarrow x=10$$

Substitute back into the original equation:

$$P(10) = -2(10)^2 + 40(10) - 100 = 100$$

So the maximum profit is £100, when 10 units are sold.

The second derivative, $\frac{d^2P}{dx^2} = -4$, is negative — confirming a maximum point.

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The Real-World Connection

This simple process mirrors how businesses use data:

• If sales grow too slowly, revenue won’t cover costs.

• If production expands too far, costs rise faster than income.

• The sweet spot — found through calculus — gives the best balance of output and efficiency.

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Skills Highlight

• Differentiating quadratic and polynomial functions

• Using the first and second derivatives to locate maxima and minima

• Interpreting results in economic and practical contexts

• Applying mathematical reasoning to real decision-making

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Why It Works in Teaching

Linking calculus to business and economics transforms it from pure theory into something purposeful. Students see that differentiation isn’t just about curves — it’s about optimisation, helping to make real-world decisions about efficiency, profit, and performance.