AP®MathsCollege BoardCalculus ABStudy GuidesUnit 5: Analytical Applications of DifferentiationGraphs of Functions & Their DerivativesOptimization Problems

Optimization Problems (College Board AP® Calculus AB): Study Guide

Written by: Jamie Wood

Reviewed by: Dan Finlay

Updated on 12 May 2026

Optimization problems

What is an optimization problem?

Recall that differentiation is about the rate of change of a function and provides a way of finding minimum and maximum values of a function
Anything that involves maximizing or minimizing a quantity can be modelled using differentiation; for example
- minimizing the cost of raw materials used in manufacturing a product
- finding the maximum height a football reaches when kicked
These are called optimization problems

How do I solve an optimization problem?

In optimization problems, variables other than x, y and f are often used
- V is often used for volume, S for surface area
- r for radius if a circle, cylinder or sphere is involved
Derivatives can still be found
- but be clear about which letter is representing the independent (x) variable
- and which letter is representing the dependent (y) variable
Problems often start by linking two connected quantities together, for example volume and surface area
- If more than one variable is involved, constraints will usually be given
  - so that the quantity being optimized can be rewritten in terms of one variable
Once the quantity being optimized is written as a function of a single variable, differentiation can be used to maximize or minimize the quantity as required

Steps to solve an optimization problem

STEP 1
Rewrite the quantity to be optimized in terms of a single variable, using any constraints given in the question
STEP 2
Differentiate and solve the derivative equal to zero to find the “x"-coordinate(s) of any critical points
STEP 3
If there is more than one critical point, or you are required to justify the nature of the critical point, differentiate again or use the candidates test
STEP 4
Use the second derivative to determine the nature of each critical point and select the maximum or minimum point as necessary or use the candidates test
STEP 5
Interpret the answer in the context of the question

Worked Example

A large flower bed is being designed as a rectangle with a semicircle on each end, as shown in the diagram below.

The total area of the bed is to be $100 π$ meters squared.

Diagram of an oval with two straight sides and semi-circular ends. A vertical line marks radius "r cm" from the center to the edge of a semi-circle.

(a) Show that the perimeter of the bed is given by the formula

$P = π (r + \frac{100}{r})$

(b) Find the value of $r$ that minimizes the perimeter, and find the minimum value of the perimeter.

Answer:

(a)

The width of the rectangle is $2 r$ meters, and its length is $L$ meters

The area consists of a semi circle, plus a rectangle, plus another semi circle

$\frac{1}{2} π r^{2} + 2 r L + \frac{1}{2} π r^{2} = 100 π$

Simplify and write $L$ in terms of $r$

$\begin{array}{rcl} π r^{2} + 2 r L & = & 100 π \\ 2 r L & = & 100 π - π r^{2} \\ L & = & \frac{50 π}{r} - \frac{π}{2} r \end{array}$

The perimeter of the flower bed consists of two semi-circular arcs, and two straight lengths

$P = π r + π r + 2 L$

Substitute in the expression for $L$ , and simplify to the desired expression for $P$

$\begin{array}{rcl} P & = & π r + π r + 2 (\frac{50 π}{r} - \frac{π}{2} r) \\ P & = & 2 π r + \frac{100 π}{r} - π r \\ P & = & π r + \frac{100 π}{r} \end{array}$

$P = π (r + \frac{100}{r})$

(b)

To find the minimum value of $P$ , we need to find the minimum point on a graph of $P$ against $r$

To do this, we need to differentiate $P$ with respect to $r$

Start by writing $P$ in terms of $r$ , using the answer from part (a)

$\begin{array}{rcl} P & = & π (r + \frac{100}{r}) \\ P & = & π r + 100 π r^{- 1} \end{array}$

Differentiate $P$ with respect to $r$

$\frac{d P}{d r} = π - 100 π r^{- 2}$

At the minimum point, the derivative will be zero (assuming the minimum is at a critical point)

$\begin{array}{rcl} π - 100 π r^{- 2} & = & 0 \\ π - \frac{100 π}{r^{2}} & = & 0 \\ 1 - \frac{100}{r^{2}} & = & 0 \\ 1 & = & \frac{100}{r^{2}} \\ r^{2} & = & 100 \\ r & = & \pm 10 \end{array}$

$r$ is a length, so -10 can be ignored

$r = 10$

Find the minimum value of the perimeter by substituting $r = 10$ into the equation for $P$

$P = π (10 + \frac{100}{10}) = 20 π$

$r = 10$ meters minimizes the perimeter

The minimum value of the perimeter will be 20π meters

(c)

To prove it is a minimum, show that the second derivative is positive (and the graph is therefore concave up) at this point

Start by finding the second derivative

$\frac{d P}{d r} = π - 100 π r^{- 2} \frac{d^{2} P}{d r^{2}} = 200 π r^{- 3}$

Substitute in $r = 10$

${\frac{d^{2} P}{d r^{2}}}_{r = 10} = 200 π {(10)}^{- 3} = \frac{200 π}{1000} = \frac{π}{5} > 0$

Note that we don't need to check endpoints here: as $r \to 0$ the perimeter function becomes unbounded, while as $r \to \infty$ the perimeter also goes to infinity

Second derivative is positive at $r = 10$ , therefore 20π meters is the minimum value for the perimeter

Worked Example

A manufacturer produces custom electronic components. The daily cost to produce x batches of components, measured in dollars, is modeled by the function $C (x) = 2 x^{3} - 21 x^{2} + 60 x + 100$ for the interval $1 \leq x \leq 6$ .

Find the absolute minimum and absolute maximum daily cost to produce the components on the closed interval $1 \leq x \leq 6$ . Justify your answers.

Answer:

Find the derivative of the function

$C' (x) = 6 x^{2} - 42 x + 60$

Set the derivative equal to zero to find critical points and solve

$C' (x) = 0 6 x^{2} - 42 x + 60 = 0 x^{2} - 7 x + 10 = 0 (x - 2) (x - 5) = 0 x = 2, x = 5$

Evaluate the function at the critical points and the endpoints

$C (1) = 2 {(1)}^{3} - 21 {(1)}^{2} + 60 (1) + 100 = 2 - 21 + 60 + 100 = 141 C (2) = 2 {(2)}^{3} - 21 {(2)}^{2} + 60 (2) + 100 = 16 - 84 + 120 + 100 = 152 C (5) = 2 {(5)}^{3} - 21 {(5)}^{2} + 60 (5) + 100 = 250 - 525 + 300 + 100 = 125 C (6) = 2 {(6)}^{3} - 21 {(6)}^{2} + 60 (6) + 100 = 432 - 756 + 360 + 100 = 136$

Justify the use of the candidates test

$C (x)$ is a polynomial, so it is continuous over $1 \leq x \leq 6$

The extreme value theorem guarantees that $C$ has a global maximum and a global minimum

These occur at the critical points or at the endpoints

State the minimum and maximum values

The absolute minimum daily cost is $125 (when 5 batches are produced)

The absolute maximum daily cost is $152 (when 2 batches are produced)

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Optimization Problems (College Board AP® Calculus AB): Study Guide

Optimization problems

What is an optimization problem?

How do I solve an optimization problem?

Steps to solve an optimization problem

Worked Example

Worked Example

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Jamie Wood

Reviewer: Dan Finlay