Optimization: Step-by-Step Guide
Optimization problems use calculus to find the maximum or minimum value of a quantity subject to a constraint. The strategy is to express the objective as a function of one variable, find its critical points, and determine which produces the extreme value within the feasible domain.
Math Racer skill guide · Calculus 1 · AP Calc AB / College
How to solve it, step by step
- Identify what to optimize and the constraint. Read the problem carefully. Define a variable for the quantity to be maximized or minimized (the objective) and write any constraint equations relating the unknowns.\[\text{Objective: } Q; \;\; \text{Constraint: } g(x, y) = 0\]
- Reduce to one variable. Use the constraint to solve for one variable in terms of the other, then substitute into the objective so it depends on a single variable.\[Q = Q(x) \text{ after substitution}\]
- Determine the feasible domain. List the values of the remaining variable that make sense physically (e.g., lengths must be nonnegative). This domain restricts the candidates.\[x \in [a, b] \text{ or open interval}\]
- Differentiate and find critical points. Compute Q'(x), set it equal to zero, and solve. Include points where Q'(x) is undefined (in the domain).\[Q'(x) = 0 \;\; \text{or undefined}\]
- Evaluate and select the extreme value. Compute Q at each critical point and at the endpoints (when the domain is closed). The largest or smallest value answers the question.\[\max/\min \;\{Q(c_i), Q(a), Q(b)\}\]
Worked example
A rancher has 100 m of fencing to enclose a rectangular pen along a river (no fence needed on the river side). Find the dimensions that maximize the enclosed area.
Step 1: Let x be the side perpendicular to the river and y the side parallel to it. Objective: maximize area A = x*y. Constraint: 2x + y = 100 (only three sides need fence).
Step 2: Solve the constraint for y: y = 100 - 2x. Substitute: A(x) = x(100 - 2x) = 100x - 2x^2.
Step 3: Domain: x must be positive and y = 100 - 2x must be positive, so 0 < x < 50.
Step 4: Differentiate and solve: A'(x) = 100 - 4x. Set A'(x) = 0: x = 25.
Step 5: Verify it is a maximum: A''(x) = -4 < 0, so x = 25 is a local (and absolute) max on (0, 50). Then y = 100 - 50 = 50, and the maximum area is A = 25*50 = 1250 m^2.
Verification: At x = 24, A = 24*52 = 1248 < 1250. At x = 26, A = 26*48 = 1248 < 1250. Confirms x = 25 maximizes.
Tips and common mistakes
- Draw a picture and label all variables before writing equations.
- Always identify the objective (what you maximize/minimize) and the constraint (what limits you) separately.
- After substituting the constraint, the objective should depend on only ONE variable.
- Determine the feasible domain BEFORE finding critical points — it eliminates spurious solutions.
- Use the second derivative test (or first derivative test) to confirm max vs. min.
- Always evaluate at endpoints when the domain is closed; the extreme can occur there.
- Check units and reasonableness — if x is a length, x must be nonnegative.
- Common pitfalls: forgetting one side of the constraint, mislabeling which dimension is x vs y, omitting endpoints.