Maxima, Minima and Optimisation: Quick Revision

A brief note on Maxima, Minima and Optimisation. Read it before a test. Try each question at the end before you open its answer.

Critical points and the first derivative test

• Critical point: a value $c$ where $f'(c) = 0$ (a stationary point) or where $f'(c)$ does not exist.
• Local extreme values: only at critical points. Not every critical point is one, as $x^3$ at $0$ shows.
• First derivative test: $f'$ from $+$ to $-$ at $c$ is a local maximum. From $-$ to $+$ is a local minimum. No change is neither.

The second derivative test and intervals

• Second derivative test: if $f'(c) = 0$, then $f''(c) < 0$ is a local maximum and $f''(c) > 0$ is a local minimum. If $f''(c) = 0$, use the first derivative test.
• Closed interval $[a, b]$: compare $f$ at the end points and at the critical points inside. The largest is the greatest value and the smallest is the least value.
• Open interval: the greatest or least value may not exist.

Optimisation steps

• Steps: draw, name the variables, write the objective in one variable using the condition, find the domain, solve $f'(x) = 0$, check with $f''$ or the end points, answer with units.
• Standard results: a rectangle of fixed perimeter has the greatest area when it is a square. A closed cylinder of fixed volume has the least surface when $h = 2r$. An open box from a square sheet of side $S$ has the greatest volume with a cut of $\dfrac{S}{6}$.

A worked example

Find the local extreme values of $f(x) = x^3 - 12x + 5$.

1. Differentiate. $f'(x) = 3x^2 - 12 = 3(x - 2)(x + 2)$, so $x = \pm 2$.
2. Second derivative. $f''(x) = 6x$. At $x = -2$, $f'' = -12 < 0$, a local maximum. At $x = 2$, $f'' = 12 > 0$, a local minimum.
3. The values. $f(-2) = -8 + 24 + 5 = 21$ and $f(2) = 8 - 24 + 5 = -11$.

Practice with new numbers

A program to try

The program finds the critical points of $f(x) = x^3 - 3x$, shows the second derivative at each, and finds the greatest and least values on $[-2, 3]$.

Mistakes to avoid

• Using the second derivative test when $f''(c) = 0$. It says nothing.
• Forgetting the end points of a closed interval.
• Forgetting a critical point where $f'$ does not exist.
• Not showing that the answer is a maximum or a minimum in an optimisation problem.

Check yourself

1. Find the critical points of $f(x) = x^2 - 4x$.

2. Classify the critical point of $f(x) = 3 - x^2$ with $f''$.

3. Find the greatest and least value of $f(x) = x^2$ on $[-1, 2]$.

4. A rectangle has a perimeter of $24\ \text{cm}$. What is its greatest area?

5. Why is $x = 0$ not an extreme point of $f(x) = x^3$?

Sources

• NCERT, Mathematics Textbook for Class XII, Part I, Chapter 6 (Application of Derivatives).
• OpenStax, Calculus Volume 1, Section 4.3 (Maxima and Minima), Section 4.5 (Derivatives and the Shape of a Graph) and Section 4.7 (Applied Optimization Problems).