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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?

  1. 4
  2. 6
  3. 2
  4. 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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