Applied derivatives

Optimization Problems

Solve constrained optimization problems by writing one-variable objective functions, finding critical points, and testing candidates.

How this page is maintained

Written for learners, checked against the sources below, and reviewed every year. Last reviewed July 22, 2026.

Short answer

Optimization uses derivatives to locate maxima or minima. Express the objective in one variable, find critical points from derivative zero or undefined points, and test candidates with endpoints or sign analysis.

  • Translate words into objective and constraint equations first.
  • Reduce objective to one variable before differentiating.
  • Compare all valid candidates, not only where derivative is zero.

Set up objective and constraint cleanly

Most optimization errors happen before calculus starts. Define the quantity to maximize or minimize, then use the constraint to eliminate one variable.

Domain limits matter. Physical dimensions cannot be negative and often have upper bounds.

Find and test critical candidates

Differentiate the one-variable objective and solve for critical points. Then test with first derivative sign changes, second derivative, or direct comparison including endpoints.

Report the final result with units and the variable values that produce it.

  • State feasible interval clearly.
  • Check endpoint values for closed intervals.
  • Reject critical points outside the domain.

Maximize rectangle area with fixed perimeter

Perimeter is 40 m. Let length be x and width be y.

  1. Constraint: 2x + 2y = 40, so y = 20 - x.
  2. Objective: A(x) = x(20 - x) = 20x - x^2, with 0 < x < 20.
  3. Derivative A'(x) = 20 - 2x = 0 gives x = 10, then y = 10.
Result: Maximum area is 100 m^2 at a 10 by 10 square.

Common mistakes

  • Differentiating before removing extra variables.
  • Ignoring endpoint checks when interval is closed.
  • Accepting algebraic critical points outside feasible domain.
  • Reporting x value without objective value and units.

Try one

Why is checking endpoints required in many maximum or minimum problems?

Because absolute extrema on a closed interval can occur at endpoints, not only at critical points.

Sources

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