Unit 5: Analytical Applications of Differentiation

Optimization Problems

Using derivatives to find absolute maximum and minimum values in context.

Advanced25 min lesson4 min readUpdated August 12, 2026Author not yet attributed

Prerequisites

The Second Derivative Test & Concavity

Finding the Best Possible Value

Every extrema tool so far has been applied to a function already given. Before reading on: for an applied problem — like enclosing the largest possible area with a fixed amount of fencing — what has to happen before any derivative can even be taken?

An optimization problem starts by translating a real constraint into a single-variable function to maximize or minimize — often using one given relationship to eliminate a second variable. Once that function exists, the first or second derivative test finds its extremum, the same way as any other critical-point analysis.

Worked Example — Maximizing Enclosed Area

A farmer has 120 meters of fencing to enclose a rectangular field along a straight river, with no fence needed along the river side. Maximize the enclosed area. Let x = length of each of the two sides perpendicular to the river, y = the side parallel to the river. The fencing constraint is 2x + y = 120, so y = 120 − 2x. Area: A(x) = xy = x(120 − 2x) = 120x − 2x², with domain 0 < x < 60 (both x and y must be positive). A'(x) = 120 − 4x; setting A'(x) = 0 gives x = 30. A''(x) = −4 < 0 everywhere, so the function is concave down throughout — confirming x = 30 is an absolute maximum, not just a local one (the only critical point of a downward-curving function on its domain is its overall peak). y = 120 − 2(30) = 60, and the maximum area is 30 × 60 = 1800 m². As x → 0⁺ or x → 60⁻, A → 0, consistent with an interior maximum.

Worked Example — Finding Absolute Extrema on a Closed Interval

Find the absolute maximum and minimum of f(x) = x³ − 6x² + 9x + 1 on [0, 5]. f'(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3), giving critical points x = 1 and x = 3, both inside [0, 5]. On a closed interval, an absolute extremum can occur at a critical point or at an endpoint, so evaluate f at all four candidates: f(0) = 1, f(1) = 1 − 6 + 9 + 1 = 5, f(3) = 27 − 54 + 27 + 1 = 1, f(5) = 125 − 150 + 45 + 1 = 21. Comparing 1, 5, 1, 21: the absolute maximum is 21, at x = 5 — an endpoint, not a critical point — and the absolute minimum is 1, tied at both x = 0 and x = 3.

Tip

On a closed interval, always test every endpoint alongside every critical point before declaring an absolute extremum — the example above shows the true maximum sitting at an endpoint entirely outside where the critical points are.

Common Mistakes

Reporting a local extremum from the first or second derivative test as though it were automatically the absolute extremum.
Local and absolute extrema are different claims — a local max is only the highest point in its immediate neighborhood; confirming it's absolute requires either a domain-wide concavity argument (as in the fencing problem) or a full candidates test against every endpoint (as in the closed-interval example).
Forgetting to restrict the domain based on the problem's physical constraints before optimizing.
In the fencing problem, x has to stay strictly between 0 and 60 for both side lengths to remain positive — optimizing over all real numbers without that restriction could return a critical point outside what the problem actually allows.

Key Takeaways

Optimization problems first translate a constraint into a single-variable function, then apply the first or second derivative test to its critical points.
A domain-wide concavity argument (a single critical point of a function concave in one direction throughout) can confirm an absolute extremum without checking endpoints.
On a closed interval, an absolute extremum can occur at an endpoint instead of a critical point — both must be checked.

Summary

This closes Unit 5: extrema, concavity, and optimization complete the derivative's picture of a function's shape and best values. Unit 6 turns to the opposite question — recovering a quantity from its rate of change, starting with approximating area under a curve.

Track your progress

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