Optimization

You’ll be able to

Translate a word problem into an objective function with a constraint
Reduce the objective to a single variable and find its extrema
Verify a candidate is the true maximum or minimum and check endpoints

Maximize or minimize a real quantity

Optimization problems ask for the largest or smallest value of some quantity — maximum area, minimum cost, shortest distance. The plan: write the objective function for the quantity to optimize, use a constraint equation to eliminate variables until the objective depends on one variable, then find critical points and test them. The setup — turning words into two equations — is the hard part.

Reduce to one variable, then optimize

Solve the constraint for one variable and substitute into the objective, so it becomes a function of a single variable on a sensible domain. Differentiate, set the derivative to zero to find critical points, and classify them. Then confirm the extremum — with the First or Second Derivative Test — and remember to check the endpoints of the domain, since a maximum on a closed interval can occur there.

Optimization strategy

objective Q(x) ← substitute constraint → optimize Q′(x) = 0 → test candidates and endpoints

A closed-interval extremum occurs at a critical point or an endpoint. Always justify which candidate wins.

Worked example

A farmer has 100 m of fence to enclose a rectangular field. What dimensions maximize the area?

1.Objective: area A = x·y. Constraint: the perimeter 2x + 2y = 100, so y = 50 − x.
2.Substitute to one variable: A(x) = x(50 − x) = 50x − x², valid for 0 < x < 50.
3.Differentiate and set to zero: A′(x) = 50 − 2x = 0, so x = 25.
4.Check it is a max: A″(x) = −2 < 0, concave down, so x = 25 is a maximum. Then y = 50 − 25 = 25.

Answer: A 25 m × 25 m square maximizes the area at 625 m². Among all rectangles of fixed perimeter, the square is always the largest — a classic result.

Tip

Name the two equations explicitly: the objective (what you are optimizing) and the constraint (the fixed condition). Solving the constraint for one variable and substituting is the step that collapses the problem to single-variable calculus.

Checkpoint

To maximize area A = x(50 − x), which equation gives the critical point?

Checkpoint

When optimizing a continuous function on a closed interval [a, b], where can the maximum occur?

On the exam

Optimization free-response points hinge on justification: after finding a critical point, state a reason it gives the extreme value (sign of f′ or f″), and on a closed domain compare it against the endpoint values. A number with no justification rarely earns full credit.

Answer the 2 checkpoints as you read.