AP Calculus BC · Unit 5: Analytical Applications of Differentiation
5.10 Introduction to Optimization Problems
What optimization problems ask
Optimization problems find the maximum or minimum value of a quantity described by a real-world scenario, subject to a constraint. Steps: (1) assign variables and write an equation for the quantity to optimize, (2) use the constraint to write it as a function of one variable, (3) find critical points using derivatives, (4) verify using the first or second derivative test.
Setting up the model
Draw a diagram when possible and label all quantities. Write the objective function (what you're optimizing, like area or cost) and the constraint equation (a relationship that limits the variables). Solve the constraint for one variable and substitute into the objective function, reducing it to a single-variable function ready for calculus techniques.
Confirming max or min, and domain
After finding critical points where f'(x) = 0 or is undefined, use the second derivative test (f''>0 = min, f''<0 = max) or check sign changes of f' to classify each point. Always restrict to the domain that makes physical sense (e.g., lengths must be positive), and check endpoints of a closed domain since extrema can occur there too.
Key terms
- Objective function
- The equation representing the quantity you want to maximize or minimize.
- Constraint equation
- A relationship between variables that restricts possible solutions, used to reduce the objective function to one variable.
- Steps to solve an optimization problem
- 1) Write objective and constraint equations, 2) reduce to one variable, 3) find critical points via derivatives, 4) classify using first/second derivative test, 5) check domain.
- Second derivative test for min/max
- f''(x) > 0 → local minimum; f''(x) < 0 → local maximum at a critical point.
- Why domain restrictions matter
- Physical quantities (length, area, time) must stay positive/realistic, so only valid critical points within the domain are considered.
Try an AP-style question
Two positive numbers have a sum of 20. What is their largest possible product?
- 100
- 96
- 20
- 400
Show the answer
A. P = x(20 − x); P' = 20 − 2x = 0 → x = 10; 10 · 10 = 100.
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