Phase 2 - Lesson 2.5

Optimization in One Variable

Finding maxima and minima with the first- and second-derivative tests, and applying them to profit maximization and utility.

⏱ 55 min● Intermediate🔗 Prereqs: 2.4

↖ Phase 2 hub

Builds on: 2.2-2.3 gave derivatives; optimization is their first big payoff.
Leads to: These ideas generalize to gradients and the KKT conditions of convex optimization (Phase 10).

Learning Objectives

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Locate critical points and classify them with the first- and second-derivative tests.
Distinguish local from global extrema and handle closed-interval endpoints.
State the roles of Fermat’s theorem and the Extreme Value Theorem in optimization.
Solve a profit-maximization problem using marginal analysis.

Key Vocabulary

Critical point: A point where f′=0 or f′ is undefined; the only candidates for interior extrema.

Fermat’s theorem: If f has a local extremum at an interior point where it is differentiable, then f′=0 there.

First-derivative test: f′ changing + to − marks a local max; − to + a local min.

Second-derivative test: At a critical point, f″<0 ⇒ local max, f″>0 ⇒ local min, f″=0 inconclusive.

Global (absolute) extremum: The largest/smallest value over the whole domain, possibly at an endpoint.

Concavity: f″>0 means convex (cup up); f″<0 means concave (cap down).

Where extrema can hide

Fermat’s theorem: at an interior local max or min of a differentiable function, the tangent is horizontal, so $f'(x)=0$. Thus interior extrema occur only at critical points. On a closed interval $[a,b]$ the Extreme Value Theorem guarantees a global max and min exist, and they sit either at a critical point or an endpoint.

Key Idea

Global-extremum recipe on [a,b]: (1) find all critical points inside, (2) evaluate f at them and at both endpoints, (3) the largest value is the global max, the smallest the global min. No calculus theorem beyond this is needed.

Classifying a critical point

Two tests decide whether a critical point is a max, a min, or neither. The first-derivative test reads the sign change of $f'$; the second-derivative test reads curvature: $f''\gt 0$ (convex) means a min, $f''\lt 0$ (concave) means a max.

Worked Example - Classify the critical points of f(x)=x³−3x

$f'(x)=3x^2-3=3(x-1)(x+1)$, so critical points at $x=\pm1$.

$f''(x)=6x$. At $x=1$, $f''=6\gt 0$ → local min $f(1)=-2$.

At $x=-1$, $f''=-6\lt 0$ → local max $f(-1)=2$.

Neither is global: $x^3-3x\to\pm\infty$, so on all of $\R$ there is no absolute extremum.

Finance link: profit maximization

A firm with revenue $R(q)$ and cost $C(q)$ maximizes profit $\pi(q)=R(q)-C(q)$. Setting $\pi'(q)=0$ gives the classic condition $R'(q)=C'(q)$ - marginal revenue equals marginal cost. The second-order condition $\pi''\lt 0$ (i.e. $R''\lt C''$) confirms it is a maximum, not a minimum.

Interactive: hunt the optimum

Common Mistakes to Avoid

Assuming every critical point is an extremum - f′=0 can be a saddle/inflection, as x³ at 0.
Reporting a local max as the global max without checking endpoints and end-behavior.
Forgetting critical points where f′ is undefined (corners), not just where f′=0.
Trusting the second-derivative test when f″=0 - it is inconclusive; fall back to the first-derivative test.

Quant Practitioner Tips

On a closed interval, always tabulate f at critical points AND endpoints before declaring a global winner.
Marginal-equals-marginal (MR=MC) is Fermat’s theorem wearing an economics hat.
Concavity (sign of f″) doubles as a max/min classifier and a convexity check for optimization theory.
If the second-derivative test stalls (f″=0), examine the sign change of f′ directly.

Knowledge Check

Q1 Easy

At an interior local maximum of a differentiable function, Fermat’s theorem guarantees:

f″=0

f′=0

f=0

f is increasing

Q2 Medium

If f′(c)=0 and f″(c)>0, then c is a:

local maximum

local minimum

saddle point

global maximum

Q3 Medium

A firm maximizes profit π=R−C. The first-order condition is:

R=C

R′=C′ (marginal revenue = marginal cost)

R′=0

C′=0

Practical Exercise

A manufacturer’s profit is $\pi(q)=-q^3+9q^2-15q-10$ (thousands of dollars) for $q\in[0,7]$ units (thousands). Find the production level that maximizes profit and the maximum profit.

▶ Show full solution

Differentiate: $\pi'(q)=-3q^2+18q-15=-3(q^2-6q+5)=-3(q-1)(q-5)$. Critical points $q=1,5$.

Second derivative $\pi''(q)=-6q+18$: at $q=5$, $\pi''=-12\lt 0$ (local max); at $q=1$, $\pi''=12\gt 0$ (local min).

Evaluate candidates and endpoints: $\pi(0)=-10,\ \pi(1)=-17,\ \pi(5)=15,\ \pi(7)=-17$. The global maximum is at $q=5$ with profit $\$15{,}000$.

After the reveal, answer for yourself: Notice the interior local min at q=1 was the WORST candidate. Always compare every critical point and endpoint numerically before crowning the maximum.

Lesson Summary

Interior extrema of a differentiable function occur only at critical points (Fermat), and on a closed interval the global extrema (guaranteed by the Extreme Value Theorem) lie among critical points and endpoints. The first- and second-derivative tests classify each candidate. In economics this is marginal analysis: profit peaks where marginal revenue equals marginal cost.

Formula Sheet Additions

Profit optimum

\[\pi'(q)=R'(q)-C'(q)=0\ \Rightarrow\ R'(q)=C'(q)\]

Marginal revenue equals marginal cost - Fermat’s theorem in economics.

Retrieval Practice

Close the lesson and answer from memory before checking. This is deliberate, effortful recall - the single highest-yield study action.

▶ Show retrieval prompts & answers

Q: Give the recipe for the global max/min of a continuous f on [a,b].
A: Find interior critical points (f′=0 or undefined), evaluate f there and at both endpoints, and pick the largest and smallest values; the Extreme Value Theorem guarantees they exist.

Q: State the second-derivative test and when it fails.
A: At a critical point c: f″(c)<0 gives a local max, f″(c)>0 a local min; if f″(c)=0 it is inconclusive and you use the first-derivative sign-change test instead.

Completion Checklist

I can explain the core ideas in my own words
I worked the derivations/examples by hand
I completed the interactive workbench(es)
I passed the knowledge check

Confidence / mastery rating

Personal notes

Source References

This lesson synthesizes and paraphrases concepts from the sources below. No copyrighted text, problem sets, or solutions are reproduced. Return to the originals for full depth.

Calculus, Vol. I (Tom Apostol, 2nd ed., 1967) foundational - Vol. I, Ch. 4 - Maxima and minima, the mean-value theorem, and the derivative tests.
A Mind for Numbers (Barbara Oakley, 2014) foundational - interleaving - Interleaving optimization problem types to train recognition of which test to apply.

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