Applications of Derivatives

Optimization Problems

Using derivatives to find maximum and minimum values in applied problems.

Advanced25 min lesson2 min readUpdated August 12, 2026Author not yet attributed

Prerequisites

The Second Derivative & Concavity

Finding the Best Possible Value

An optimization problem first translates a constraint into a single-variable function, then applies a derivative test to its critical points.

Worked Example — Maximizing a Rectangle's Area with a Fixed Perimeter

200 meters of fencing must enclose a rectangular field on all four sides. Maximize the enclosed area. Let x and y be the side lengths: 2x + 2y = 200, so y = 100 − x. Area: A(x) = xy = x(100 − x) = 100x − x², domain 0 < x < 100. A'(x) = 100 − 2x; setting A'(x) = 0 gives x = 50. A''(x) = −2 everywhere, so A is concave down throughout — confirming x = 50 is an absolute maximum. y = 100 − 50 = 50, so the optimal shape is a 50-by-50 square, with maximum area 50 × 50 = 2500 m².

Tip

A domain-wide negative second derivative (concave down everywhere) confirms a single critical point is an absolute maximum, without needing to check endpoints separately.

Common Mistakes

Optimizing a two-variable expression directly, without first using the constraint to eliminate one variable.
A derivative test requires a single-variable function — the constraint (here, the fixed perimeter) is exactly what lets y be rewritten in terms of x before differentiating.

Key Takeaways

Optimization problems translate a constraint into a single-variable function, then apply a derivative test to its critical points.
A domain-wide concavity argument can confirm an absolute extremum without checking endpoints.

Summary

The next lesson uses derivatives to connect multiple changing quantities at a shared instant.

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