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Recipe

Solve an optimization problem

Use when: a word problem asks for the largest or smallest possible value of some quantity, subject to a constraint

Goal: the input value that actually maximizes or minimizes the quantity, plus that optimal value itself

Taught in Optimization: the general method

Quick recipe

  1. Identify the quantity to optimize and the constraint
  2. Reduce to a single variable
  3. Find the critical points
  4. Confirm it's actually a max or min
  5. Check the domain's endpoints too
  1. 1

    Name the quantity you're maximizing or minimizing, and write down the relationship (the constraint) tying its variables together.

    Example

    Splitting 20 into x+y=20 to maximize the product P=xy: the quantity is P, the constraint is x+y=20.

    Why: a clearly named quantity and constraint are what make every later algebra step unambiguous. Most optimization mistakes trace back to a vague setup here.

  2. 2

    Use the constraint to eliminate one variable, rewriting the quantity as a function of one remaining variable.

    Example

    From y=20-x: P(x)=x(20-x)=20x-x^2.

    Why: the tools this course has for finding a max or min (critical points, sign charts) all apply to a function of ONE variable.

  3. 3

    Differentiate, set the derivative equal to 0 (and check where it's undefined), and solve.

    Example

    P'(x)=20-2x; setting this to 0 gives x=10.

    Why: a maximum or minimum on the interior of the domain can only happen at a critical point. This narrows down every possible candidate.

  4. 4

    Use the First or Second Derivative Test to classify each critical point.

    Example

    P''(x)=-2<0 everywhere, so x=10 gives a maximum by the Second Derivative Test.

    Why: a critical point is only a CANDIDATE. Without this check it could just as easily be a local min, or neither.

  5. 5

    If the variable is restricted to a closed interval, compare the critical point's value against the value at each endpoint.

    • the domain is all reals (or an open, unbounded interval) — the confirmed critical point from Step 4 is the answer.
    • the domain is a closed interval [a,b] — evaluate the quantity at a, at b, and at every interior critical point, then compare all three.

    Why: the Extreme Value Theorem only guarantees a max and min exist on a closed interval — and either one might sit at an endpoint the derivative test never examines.

Common mistakes

  • Optimizing a two-variable expression without first reducing it to one variable.
  • Finding a critical point but never checking whether it's actually a max, a min, or neither.
  • Ignoring domain restrictions that come from the problem's own physical setup (like a length that can't be negative).
  • Reporting the critical NUMBER as the answer when the problem asked for the optimized quantity's VALUE.