AP Calculus BC · Unit 5: Analytical Applications of Differentiation
5.11 Solving Optimization Problems
What optimization problems ask
Optimization problems ask you to find the maximum or minimum value of a quantity described by a real-world situation (area, volume, cost, distance). You write a function for the quantity, restrict its domain based on physical constraints, then use calculus to find where it's largest or smallest — usually on a closed interval.
The standard procedure
1) Draw a diagram and define variables. 2) Write a primary equation for the quantity to optimize and a constraint equation relating variables. 3) Substitute to get a single-variable function. 4) Find critical points by setting the derivative equal to 0. 5) Use the first or second derivative test (or check endpoints) to confirm max/min.
Justifying the answer
Always justify that your critical point gives an absolute extremum, not just a local one. On a closed interval, compare the critical point's value to the endpoint values (Extreme Value Theorem). On an open/unbounded interval, use the first derivative test (sign change) or second derivative test (concavity) to justify.
Key terms
- Primary equation
- The equation representing the quantity you want to maximize or minimize.
- Constraint equation
- An equation relating the variables in the problem, used to reduce the primary equation to one variable.
- Extreme Value Theorem (optimization use)
- On a closed interval, a continuous function's absolute max/min occurs at a critical point or an endpoint.
- First derivative test
- If f' changes sign from + to − at a critical point, it's a local max; − to +, a local min.
- Steps for solving optimization problems
- 1) Diagram & define variables. 2) Write primary & constraint equations. 3) Substitute to one variable. 4) Find critical points. 5) Justify max/min using derivative tests or endpoints.
Try an AP-style question
Equal squares of side x are cut from the corners of a 12 × 12 sheet to make an open box. Which x maximizes the volume?
- 4
- 6
- 2
- 3
Show the answer
C. V = x(12 − 2x)²; V' = (12 − 2x)(12 − 6x) = 0 → x = 2 (x = 6 gives zero volume).
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