Applications of Derivatives: Maxima and Minima

Finding the highest or lowest value of a function is one of the most useful things calculus can do. Engineers use it to cut material costs, economists use it to maximise profit, and you will use it in almost every FSc, A Level and first-year university calculus paper. This guide shows you exactly how derivatives locate maxima and minima, step by step.

What Are Maxima and Minima?

A local maximum is a point where the function is higher than at all nearby points, like the top of a hill. A local minimum is a point lower than everything around it, like the bottom of a valley. Together they are called extreme values or turning points.

There is also a difference between local and absolute extremes:

• A local (relative) extreme is the highest or lowest point in its neighbourhood.
• An absolute (global) extreme is the highest or lowest value over the whole domain or a given interval.

Why the Derivative Is Zero at a Turning Point

At the top of a smooth hill or bottom of a smooth valley, the tangent line is horizontal. A horizontal tangent has slope zero, so at a turning point:

$$ f'(x) = 0 $$

Points where \( f'(x) = 0 \) or where \( f'(x) \) does not exist are called critical points. Every local extreme of a function happens at a critical point, but not every critical point is an extreme. For example, \( f(x) = x^3 \) has \( f'(0) = 0 \), yet \( x = 0 \) is a point of inflection, not a maximum or minimum.

Two Tests for Classifying Critical Points

The First Derivative Test

Check the sign of \( f'(x) \) just before and just after the critical point:

• \( f' \) changes from positive to negative: local maximum.
• \( f' \) changes from negative to positive: local minimum.
• \( f' \) does not change sign: neither (often an inflection point).

The Second Derivative Test

Substitute the critical point into \( f''(x) \):

• \( f''(c) < 0 \): the curve is concave down, so \( x = c \) is a local maximum.
• \( f''(c) > 0 \): the curve is concave up, so \( x = c \) is a local minimum.
• \( f''(c) = 0 \): the test is inconclusive; use the first derivative test instead.

The second derivative test is usually faster, which is why most textbook solutions use it.

Worked Example 1: Finding and Classifying Turning Points

Find the local maximum and minimum of \( f(x) = x^3 - 6x^2 + 9x + 1 \).

Step 1: Differentiate.

$$ f'(x) = 3x^2 - 12x + 9 $$

Step 2: Set \( f'(x) = 0 \) and solve.

$$ 3x^2 - 12x + 9 = 0 \quad\Rightarrow\quad 3(x^2 - 4x + 3) = 0 \quad\Rightarrow\quad 3(x - 1)(x - 3) = 0 $$

So the critical points are \( x = 1 \) and \( x = 3 \).

Step 3: Find the second derivative.

$$ f''(x) = 6x - 12 $$

Step 4: Classify each point.

• At \( x = 1 \): \( f''(1) = 6 - 12 = -6 < 0 \), so this is a local maximum.
• At \( x = 3 \): \( f''(3) = 18 - 12 = 6 > 0 \), so this is a local minimum.

Step 5: Find the function values.

$$ f(1) = 1 - 6 + 9 + 1 = 5, \qquad f(3) = 27 - 54 + 27 + 1 = 1 $$

Answer: local maximum at \( (1, 5) \) and local minimum at \( (3, 1) \).

Worked Example 2: An Optimisation Problem

A square sheet of cardboard with side 12 cm has equal squares of side \( x \) cm cut from each corner. The sides are folded up to make an open box. Find the value of \( x \) that gives the largest volume.

Step 1: Write the quantity to maximise.

After cutting, the base is \( (12 - 2x) \) by \( (12 - 2x) \) and the height is \( x \):

$$ V(x) = x(12 - 2x)^2, \qquad 0 < x < 6 $$

Step 2: Differentiate using the product rule.

$$ V'(x) = (12 - 2x)^2 + x \cdot 2(12 - 2x)(-2) $$

Factor out \( (12 - 2x) \):

$$ V'(x) = (12 - 2x)\big[(12 - 2x) - 4x\big] = (12 - 2x)(12 - 6x) $$

Step 3: Solve \( V'(x) = 0 \).

This gives \( x = 6 \) or \( x = 2 \). Since \( x = 6 \) makes the base zero (no box), we take \( x = 2 \).

Step 4: Confirm it is a maximum.

For \( x \) slightly less than 2 (say \( x = 1 \)): \( V'(1) = (10)(6) = 60 > 0 \). For \( x \) slightly more (say \( x = 3 \)): \( V'(3) = (6)(-6) = -36 < 0 \). The sign changes from positive to negative, so \( x = 2 \) is a maximum.

Step 5: Find the maximum volume.

$$ V(2) = 2(12 - 4)^2 = 2 \times 64 = 128 \text{ cm}^3 $$

Worked Example 3: Absolute Extremes on a Closed Interval

Find the absolute maximum and minimum of \( f(x) = x^3 - 3x \) on \( [0, 2] \).

On a closed interval, the absolute extremes occur either at critical points inside the interval or at the endpoints.

Step 1: \( f'(x) = 3x^2 - 3 = 0 \Rightarrow x = \pm 1 \). Only \( x = 1 \) lies in \( [0, 2] \).

Step 2: Evaluate at the critical point and both endpoints.

$$ f(0) = 0, \qquad f(1) = 1 - 3 = -2, \qquad f(2) = 8 - 6 = 2 $$

Answer: absolute maximum is 2 at \( x = 2 \); absolute minimum is \( -2 \) at \( x = 1 \).

A Simple Method for Optimisation Word Problems

1. Draw a diagram and label the variables.
2. Write a formula for the quantity you want to maximise or minimise.
3. Use the given condition (fixed perimeter, area, cost and so on) to reduce it to one variable.
4. State the allowed domain.
5. Differentiate, set equal to zero and solve.
6. Confirm maximum or minimum with a derivative test.
7. Answer the actual question, with units.

Common Mistakes

• Forgetting to check endpoints on a closed interval. The absolute extreme is often at an endpoint.
• Giving only the \( x \)-value. If the question asks for the maximum value, substitute back to find \( f(x) \).
• Ignoring the domain. In Example 2, \( x = 6 \) is a solution of \( V'(x) = 0 \) but makes no physical sense.
• Assuming \( f'(x) = 0 \) always means a turning point. Always classify the point.
• Sign errors in the product rule, especially with expressions like \( (12 - 2x)^2 \).

Key Takeaways

• Local maxima and minima occur at critical points, where \( f'(x) = 0 \) or \( f'(x) \) is undefined.
• If \( f''(c) < 0 \) the point is a maximum; if \( f''(c) > 0 \) it is a minimum; if \( f''(c) = 0 \), use the first derivative test.
• For absolute extremes on \( [a, b] \), compare the function values at critical points and both endpoints.
• Optimisation problems become routine once you reduce the quantity to a single variable and state its domain.

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