TOPIC 3 / 4

Derivative Tests & Optimization

Useful preparation: Derivatives, Chain Rule & Implicit Change

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

Optimization combines a model, a feasible domain, and comparison. Derivatives identify candidates. Explain the objective, find all relevant candidates, and verify which attains the best allowed value.

Local and absolute extrema: A local extremum compares nearby feasible inputs; an absolute extremum compares the entire specified domain.
Critical number: An interior domain input where the derivative is zero or does not exist. Such an input is a candidate, not a guaranteed extremum.
Feasible domain: All inputs satisfying the problem's mathematical and contextual restrictions, including any boundary points that are genuinely allowed.

Conditions before calculation

A continuous real function on a closed bounded interval attains an absolute maximum and minimum. Open or unbounded domains need separate existence analysis.
An interior differentiable local extremum has zero derivative, but the converse fails. Nondifferentiable points and endpoints need attention too.
The second-derivative test applies at a stationary point when the second derivative is continuous nearby; a zero second derivative leaves this test inconclusive.

Why the method works

1. Reduce the model without losing feasibility

Use constraints to reduce the objective to one variable, retaining its feasible interval. Physically impossible stationary points cannot solve the problem.

x+y=s,\quad x,y\ge0\ \Longrightarrow\ A(x)=x(s-x),\quad 0\le x\le s

2. Classify local behavior

A positive-to-negative derivative change means rise then fall; the reverse means fall then rise. At a stationary point, nonzero curvature supplies the second-derivative test.

f'(c)=0:\qquad f''(c)>0\Rightarrow\text{local minimum},\quad f''(c)<0\Rightarrow\text{local maximum}

3. Make the global comparison

Evaluate critical numbers and endpoints, comparing function values rather than derivative values. On an open domain, a limiting value may never be attained.

\{f(a),f(b)\}\cup\{f(c):c\in(a,b),\ f'(c)=0\text{ or }f'(c)\text{ undefined}\}

Worked examples

EXAMPLE 1

A rectangle uses 36 meters of boundary material. Find the largest area.

Let x and y be side lengths. The perimeter constraint leaves one free variable; zero lengths describe degenerate boundary cases used for comparison.
$2x+2y=36,\qquad y=18-x,\qquad A(x)=18x-x^2,\quad 0\le x\le18$
Set the derivative to zero. Its sign is positive before 9 and negative after 9, so area increases then decreases.
$A'(x)=18-2x,\qquad A'(x)=0\iff x=9$
Compare the candidate with both endpoints, and recover the other dimension.
$A(0)=A(18)=0,\qquad A(9)=81,\qquad y=9$

Conclusion: A square with side length 9 meters attains the maximum area of 81 square meters.

EXAMPLE 2

Find all locations of the absolute extrema of the cubic on the displayed closed interval.

Continuity guarantees extrema, so gather interior stationary points and endpoints.
$f(x)=x^3-3x,\quad -2\le x\le3;\qquad f'(x)=3(x^2-1)=0\iff x=\pm1$
Evaluate every candidate. This step captures both boundary effects and ties.
$f(-2)=-2,\qquad f(-1)=2,\qquad f(1)=-2,\qquad f(3)=18$
Compare the four values. The interior local maximum is smaller than the right endpoint value; the minimum occurs at two distinct inputs.
$\max f=18\text{ at }x=3,\qquad \min f=-2\text{ at }x=-2,1$

Conclusion: A local maximum need not be the absolute maximum, and an absolute extremum need not have a unique location.

Pitfalls & counterexamples

Tempting claim: Solving for zero derivative completes an optimization problem.

Correction: Stationarity only produces candidates; domain restrictions, nondifferentiable points, endpoints, and attainment must still be checked.

Test the claim: The cubing function has zero derivative at zero but increases through that point, so there is no local extremum there.

Check your understanding

Does the function equal to its input have an absolute maximum on the open interval from zero to one?

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Compare your reasoning

No.

Its values approach one but never reach it; any allowed input can be increased while remaining inside the interval.

Connect this idea

The same candidate-and-comparison logic extends to constrained multivariable optimization. Existence theorems explain why checking finitely many candidates can certify a global answer.

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Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.