01 Key Concepts
General Strategy
1) Write a formula for the quantity to optimize. 2) Use a constraint to reduce it to one variable. 3) Differentiate, set equal to zero, and solve. 4) Confirm it's a max or min.
First Derivative Test
If f' changes from positive to negative at a critical point, it's a local max; from negative to positive, a local min.
Second Derivative Test
If f''(c) > 0 at a critical point c, it's a local minimum. If f''(c) < 0, it's a local maximum.
Constraints
A limiting condition (like fixed perimeter or fixed cost) used to express one variable in terms of another, reducing the problem to a single variable.
Endpoints Matter
On a closed interval, always also check the endpoints — the true maximum or minimum might occur there instead of at a critical point.
02 Solved Examples
- Let width = x, length = y. Constraint: 2x + 2y = 40, so y = 20 - x.
- Area A = x*y = x(20-x) = 20x - x^2.
- Differentiate: A' = 20 - 2x. Set to 0: x = 10.
- Then y = 20 - 10 = 10.
- Let the numbers be x and 20-x. Product P = x(20-x) = 20x - x^2.
- Differentiate: P' = 20 - 2x. Set to 0: x = 10.
- The other number is 20 - 10 = 10.
- Differentiate: C'(x) = 2x - 20. Set to 0: x = 10.
- Check second derivative: C''(x) = 2 > 0, confirming a minimum.
- Minimum cost = C(10) = 100 - 200 + 150.
03 Practice Questions
📄 Optimization — Downloadable Worksheet
10 questions with a full answer key. Grab the PDF to print, or try the interactive version in your browser.