Optimization

Use derivatives to find and interpret maximum and minimum values.

Optimization means finding the best value: largest profit, smallest cost, maximum area, minimum error, shortest time, or highest efficiency. Derivatives help because smooth functions often have their best interior values where the rate of change is zero.

Learning objectives

After this lesson you should be able to:

  • distinguish local and absolute maxima and minima;
  • find critical points;
  • classify candidates with derivative signs;
  • use the second derivative test in simple cases;
  • check endpoints on closed intervals;
  • solve basic applied optimization problems step by step;
  • use SymPy and edumath to compute and compare candidates.

Vocabulary

  • Local maximum: higher than nearby values.
  • Local minimum: lower than nearby values.
  • Absolute maximum: highest value on the entire domain being considered.
  • Absolute minimum: lowest value on the entire domain being considered.
  • Critical point: a point in the domain where \(f'(x)=0\) or where \(f'(x)\) is undefined.
  • Candidate: a point that might produce a maximum or minimum.

Optimization roadmap

  1. Read the goal. Are you maximizing or minimizing? Which quantity?
  2. Find the domain. Are there endpoints, physical restrictions, or constraints?
  3. Differentiate. Compute \(f'(x)\).
  4. Find critical points. Solve \(f'(x)=0\) and consider where \(f'\) is undefined while \(f\) exists.
  5. Build a candidate list. Include endpoints if the domain is closed.
  6. Classify or compare. Use derivative signs, second derivative, or a table of function values.
  7. Interpret. State both the input and output, with units when needed.

Critical points

Walkthrough: find a critical point

Find the critical point of

\[ f(x)=x^2-6x+10. \]

  1. Differentiate. \[ f'(x)=2x-6. \]
  2. Set derivative equal to zero. \[ 2x-6=0. \]
  3. Solve. \[ 2x=6, \qquad x=3. \]
  4. Check domain. The function is a polynomial, so \(x=3\) is in the domain.
  5. Interpret. The only critical point is \(x=3\).

First derivative test

The first derivative tells whether the function is increasing or decreasing.

  • If \(f'\) changes from positive to negative, the function changes from increasing to decreasing: local maximum.
  • If \(f'\) changes from negative to positive, the function changes from decreasing to increasing: local minimum.
  • If \(f'\) does not change sign, the critical point may not be an extremum.

Walkthrough: sign chart

Classify the critical point of \(f(x)=x^2-6x+10\).

  1. We found \(f'(x)=2x-6\) and critical point \(x=3\).
  2. Choose test points around 3.
    • At \(x=2\), \(f'(2)=4-6=-2\), so the function is decreasing.
    • At \(x=4\), \(f'(4)=8-6=2\), so the function is increasing.
  3. The sign changes from negative to positive.
  4. Therefore \(x=3\) is a local minimum.
  5. The minimum value is \[ f(3)=9-18+10=1. \]

Second derivative test

The second derivative measures concavity.

  • If \(f''(c)>0\) and \(f'(c)=0\), the graph is concave up at \(c\): local minimum.
  • If \(f''(c)<0\) and \(f'(c)=0\), the graph is concave down at \(c\): local maximum.
  • If \(f''(c)=0\), the test is inconclusive.

For \(f(x)=x^2-6x+10\), we have \(f''(x)=2>0\), so the critical point at \(x=3\) is a local minimum.

Absolute extrema on closed intervals

On a closed interval \([a,b]\), absolute maximum and minimum values can occur at critical points or endpoints.

Walkthrough: closed interval candidate table

Find the absolute maximum and minimum of

\[ f(x)=x^3-3x \]

on \([-2,2]\).

  1. Differentiate. \[ f'(x)=3x^2-3=3(x^2-1). \]

  2. Solve \(f'(x)=0\). \[ 3(x^2-1)=0 \Rightarrow x^2=1 \Rightarrow x=-1,1. \]

  3. Build candidate list. Include endpoints and critical points: \[ x=-2,-1,1,2. \]

  4. Evaluate original function.

    Candidate Source \(f(x)\)
    -2 endpoint \(-8+6=-2\)
    -1 critical point \(-1+3=2\)
    1 critical point \(1-3=-2\)
    2 endpoint \(8-6=2\)

  5. Compare values. The absolute maximum value is 2 at \(x=-1\) and \(x=2\). The absolute minimum value is -2 at \(x=-2\) and \(x=1\).

WarningCommon mistake

Do not compare derivative values when looking for absolute maxima and minima. After finding candidates, evaluate the original function at each candidate.

Applied optimization workflow

Word problems need modeling before calculus.

  1. Draw or describe the situation.
  2. Define variables and units.
  3. Write the objective quantity.
  4. Use constraints to rewrite the objective with one variable.
  5. State the feasible domain.
  6. Differentiate and find candidates.
  7. Check endpoints and context.
  8. Interpret the answer in a sentence.

Walkthrough: maximum area with fixed perimeter

A rectangle has perimeter 40 meters. What dimensions maximize its area?

  1. Define variables. Let length be \(L\) meters and width be \(W\) meters.
  2. Write the objective. Area is \[ A=LW. \]
  3. Use the constraint. Perimeter is \[ 2L+2W=40. \] Divide by 2: \[ L+W=20. \] Solve for \(L\): \[ L=20-W. \]
  4. Write area as one-variable function. \[ A(W)=W(20-W)=20W-W^2. \]
  5. State domain. Width must be between 0 and 20.
  6. Differentiate. \[ A'(W)=20-2W. \]
  7. Find critical point. \[ 20-2W=0 \Rightarrow W=10. \]
  8. Find length. \[ L=20-10=10. \]
  9. Check endpoints. At \(W=0\) or \(W=20\), area is 0. At \(W=10\), area is 100.
  10. Interpret. The maximum-area rectangle is a \(10\) m by \(10\) m square.

Guided exercises

  1. Find the critical point of \(x^2-8x+3\).
  2. Classify the critical point of \(x^2\) at \(x=0\).
  3. If \(f'\) changes from positive to negative, what kind of point is suggested?
  4. Why do we check endpoints on \([a,b]\)?
  5. A revenue model is \(R(x)=-x^2+10x\). Find the production level that maximizes revenue.
  1. \(f'(x)=2x-8\), so \(2x-8=0\) and \(x=4\).
  2. \(x=0\) is a minimum because \(x^2\) decreases before 0 and increases after 0.
  3. A local maximum.
  4. The absolute best value can occur at the boundary of the interval.
  5. \(R'(x)=-2x+10\). Set to zero: \(x=5\).

Guessing game checkpoint

Using this lesson with edumath and SymPy

Use SymPy to compute candidates, but still build the interpretation yourself.

import sympy as sp

x = sp.symbols("x")
f = x**2 - 6*x + 10
critical = sp.solve(sp.Eq(sp.diff(f, x), 0), x)
critical
[3]
[(c, f.subs(x, c)) for c in critical]
[(3, 1)]
from edumath.calculus import optimization_candidate_table, validate_critical_points

optimization_candidate_table("x**3 - 3*x", domain=(-2, 2))
(OptimizationCandidate(location=-1, value=2, source='critical point'),
 OptimizationCandidate(location=1, value=-2, source='critical point'),
 OptimizationCandidate(location=-2, value=-2, source='endpoint'),
 OptimizationCandidate(location=2, value=2, source='endpoint'))
validate_critical_points("3", "x**2 - 6*x + 10").message
'Correct.'

Further reading