Short on time? This page is the recap. For explanations, diagrams and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Recap
- An optimization problem asks for the largest or smallest value of one quantity on an interval.
- Every problem has the same skeleton: quantity to optimise, constraint, interval.
- The constraint lets you write the quantity as a function of one variable. Only then differentiate.
- At an interior maximum or minimum of a differentiable function, the derivative is 0. So the search starts with critical points.
- Then prove which critical point gives the extreme value, and answer exactly what was asked (value, location or both).
Key relationships
| Situation | How to prove the extreme value |
|---|---|
| Continuous function on a closed interval [a, b] | Candidates test: evaluate at every critical point and both endpoints; largest is the maximum, smallest is the minimum |
| Open interval, exactly one critical point c | First derivative test (sign change of f′ at c) or second derivative test (sign of f″(c)), then “only critical point, so absolute” |
| Open interval, several critical points | Compare values, and check behaviour near the ends of the interval |
The method in eight steps: name the quantity → diagram and letters → formula → use constraint → interval → critical points → prove → answer.
Assumptions behind the method
- The function is continuous on the interval (true for polynomials and for rational functions away from zeros of the denominator).
- The interval comes from the situation: lengths positive, times within the model’s range.
- The single-critical-point argument needs the function to be continuous on an interval with only one critical point.
Mistakes to avoid
- Differentiating a formula that still has two variables.
- Stopping at f′(x) = 0 without showing it is a maximum (or minimum).
- Forgetting endpoints on a closed interval.
- Calling a relative extremum absolute without the single-critical-point reason or a comparison.
- Answering with the location when the question asks for the value.
- Keeping a critical point outside the interval, such as a negative length.
Quick self-check
- Find the maximum and minimum values of f(x) = x³ − 12x on [0, 3]. (f′ = 3x² − 12 = 0 at x = 2 in the interval. f(0) = 0, f(2) = −16, f(3) = −9. Maximum 0 at x = 0; minimum −16 at x = 2.)
- Two numbers add to 20. What is the largest possible product? (P = x(20 − x), P′ = 20 − 2x = 0 at x = 10, P″ = −2 < 0, only critical point: maximum product 100.)
- On an open interval, g′(c) = 0 and g″(c) < 0. Why is that not yet enough to say g has its absolute maximum at c? (It only shows a relative maximum. You also need c to be the only critical point in the interval, or a comparison of values.)
Next: practice questions.