Calculus · Step-by-step guide

Optimization Problems

Solve calculus optimization problems by building an objective function, finding critical points, and verifying extrema.

By Tyler Blovat··9 min read
The short answer

Start with the central idea

Choose the quantity to maximize or minimize, use constraints to write it as a one-variable function, find critical points from the derivative, and compare feasible candidates including endpoints.

01

How it works

Build the method from meaning before memorizing the moves.

These are the decisions that make the procedure reliable. Read them once, then look for each idea in the worked examples below.

1

Separate objective and constraint

The objective is what changes toward a best value; constraints describe what must remain true.

2

Reduce to one variable

Use the constraint to eliminate extra variables before differentiating.

3

Verify the best candidate

A critical point is only a candidate. Use sign changes, a second derivative, or endpoint comparison.

02

See the structure

A picture makes the relationships easier to remember.

tangent line
The derivative at the marked point is the slope of the highlighted tangent line.
03

Worked examples

Follow the reason for each line, then try to reproduce it without looking.

Example 1Rectangle perimeter

Maximize area with perimeter 40.

  1. 2l+2w=40\displaystyle 2l+2w=40, so w=20−l\displaystyle w=20-l.
  2. A(l)=l(20−l)\displaystyle A(l)=l(20-l).
  3. A′=20−2l=0\displaystyle A'=20-2l=0, so l = 10 and w = 10.
AnswerMaximum area 100
Example 2Open-top box

Cut squares x from a 12-by-12 sheet.

  1. V=x(12−2x)2\displaystyle V=x(12-2x)^2.
  2. Restrict 0<x<6\displaystyle 0<x<6.
  3. Find critical points and compare feasible values.
AnswerUse the feasible critical point x=2\displaystyle x=2
Example 3Closest point

Find the point on y = x² nearest (0,3).

  1. Minimize squared distance D2=x2+(x2−3)2\displaystyle D^2=x^2+(x^2-3)^2.
  2. Differentiate and solve critical points.
  3. Compare their squared distances.
AnswerMinimum occurs at x=±5/2\displaystyle x=\pm\sqrt{5/2}
04

Common mistakes—and how to avoid them

Accuracy improves fastest when you know what to check.

  • Differentiating before reducing to one variable.

  • Ignoring the physical domain.

  • Assuming every critical point is the requested maximum or minimum.

05

Turn the explanation into a skill

Reading creates recognition. Independent practice creates recall.

Free printable worksheet · Answer key included

Applications of Derivatives

Practice objective functions, constraints, extrema, and derivative applications.

Preview the worksheet

A strong practice loop: solve one example with the guide open, solve a similar problem without it, explain the method aloud, and return the next day for a short mixed review.

06

Practice problems with worked solutions

Solve each problem on paper, use the hint only if needed, then compare every step.

1

Two positive numbers sum to 20. Maximize their product.

Show hint

Write one number as 20 − x.

Show worked solution
  1. P=x(20−x)=20x−x2\displaystyle P=x(20-x)=20x-x^2.
  2. P′=20−2x=0\displaystyle P'=20-2x=0, so x=10\displaystyle x=10.
  3. The other number is 10.

Answer: Maximum product 100 at 10 and 10.

2

A rectangle has perimeter 40. Maximize its area.

Show hint

Use y=20−x\displaystyle y=20-x.

Show worked solution
  1. A=x(20−x)\displaystyle A=x(20-x).
  2. A′=20−2x=0\displaystyle A'=20-2x=0, so x=10\displaystyle x=10.
  3. Then y=10\displaystyle y=10.

Answer: Maximum area 100 square units.

3

Minimize f(x)=x2−8x+20\displaystyle f(x)=x^2-8x+20.

Show hint

Find the critical point.

Show worked solution
  1. f′(x)=2x−8\displaystyle f'(x)=2x-8.
  2. 2x−8=0\displaystyle 2x-8=0 gives x=4\displaystyle x=4.
  3. f(4)=16−32+20=4\displaystyle f(4)=16-32+20=4.

Answer: Minimum value 4 at x = 4.

4

An open-top box is made from a 12-by-12 sheet by cutting x-inch corners. Write its volume function.

Show hint

Dimensions become x by 12−2x by 12−2x.

Show worked solution
  1. The height is x.
  2. Both base dimensions are 12−2x\displaystyle 12-2x.
  3. Multiply the three dimensions.

Answer: V(x)=x(12−2x)2\displaystyle V(x)=x(12-2x)^2, 0<x<6\displaystyle 0<x<6.

5

Find the point on y=x2\displaystyle y=x^2 closest to (0,3)\displaystyle (0,3). Set up the quantity to minimize.

Show hint

Minimize squared distance.

Show worked solution
  1. D2=(x−0)2+(x2−3)2\displaystyle D^2=(x-0)^2+(x^2-3)^2.
  2. Expand: D2=x4−5x2+9\displaystyle D^2=x^4-5x^2+9.
  3. Differentiate: 4x3−10x=2x(2x2−5)\displaystyle 4x^3-10x=2x(2x^2-5).

Answer: Candidates x=0\displaystyle x=0 and x=±5/2\displaystyle x=\pm\sqrt{5/2}; the minima occur at x=±5/2\displaystyle x=\pm\sqrt{5/2}.

07

Frequently asked questions

Quick answers before you move on.

What should I remember first?

Choose the quantity to maximize or minimize, use constraints to write it as a one-variable function, find critical points from the derivative, and compare feasible candidates including endpoints.

How do I know whether I understand this topic?

You should be able to name the method, explain why each step is valid, complete a new example without copying, and check whether your answer is reasonable.

What should I do if I keep making the same mistake?

Write the error as a specific checkpoint—for example, “I will identify the hypotenuse before substituting.” Then use that checkpoint on three short problems rather than repeating a full page without feedback.

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