5.5 Optimization

Achievable AP Calculus AB

5. Analytical uses

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.

Optimization

5 min read

Font

Discuss

Feedback

What you’ll learn

How to solve optimization problems with a 4-step strategy
How to apply the EVT to optimization problems
How to optimize motion functions

A common real-world application of derivative tests is optimization, which means finding the maximum or minimum value of a quantity while meeting certain constraints.

Example 1: General strategy

The following example will be used to show the step-by-step process:

Let $y = 12 - x^{2}$ for $x > 0$ .

Find the value of $x$ that maximizes the product of $x$ and $y$ .

Solution

Step 1: Identify the objective function + constraint equation

The objective function is the quantity to maximize or minimize. We want to maximize the product, $P$ :

$P = x y$

The constraint equation is the rule that limits your variables. Here, the constraint is given as a function of $x$ :

$y = 12 - x^{2}$

Step 2: Rewrite the objective function in terms of a single variable.

Substitute the constraint into the objective function.

$P (x) = x (12 - x^{2}) = 12 x - x^{3}$

Step 3: Find critical points:

To find where the product is maximized, find the critical points by taking the derivative and setting it equal to zero ( $P^{'} (x) = 0$ ). Differentiating,

$P^{'} (x) = 12 - 3 x^{2}$

Then solving for the critical points:

$12 - 3 x^{2} 124 x = 0 = 3 x^{2} = x^{2} = \pm 2$

Since the problem states $x > 0$ , our only critical point is $x = 2$ .

Step 4: Justify extrema of the objective function

Use either the 1st or 2nd derivative test to classify the critical point as extrema. Here, we’ll use the 2nd derivative to check that $x = 2$ yields a maximum. Differentiating again,

$P^{''} (x) = - 6 x$

Evaluating,

$P^{''} (2) = - 6 (2) = - 12$

Because $P^{''} (2) < 0$ , the function is concave down at this point, which proves that $x = 2$ produces a relative maximum.

Sidenote

Not related rates

Look for keywords that distinguish the two, such as

Optimization: Maximum, minimum, largest, or least
Related rates: Increasing at a rate, decreasing, or changing with respect to time

Even though the two are different, time can still occasionally act as the independent variable. If a problem asks you to find the exact moment a function is maximized or minimized (rather than finding a rate of change), it is an optimization problem.

Because these time-based contexts usually happen over a specific, closed interval of time, you must use the Extreme value theorem (EVT) and compare the critical points with the endpoints of the interval to find the absolute maximum or minimum.

Example 2: Closed interval optimization

The rate at which a processing plant refines oil is modeled by the function $R (t) = t^{2} - 6 t + 10$ barrels per hour, where $t$ is measured in hours for $0 \leq t \leq 4$ . At what time $t$ is the refinery processing oil at the absolute slowest rate?

Solution

(spoiler)

The keyword “absolute” signals use of the EVT, while “slowest” frames it as an optimization problem to find the minimum of the function.

1. Identify the objective function + constraint:

We want to minimize the refining rate, $R (t)$ .

There is no constraint equation since the function is already in terms of a single variable, $t$ .

2. Find the critical points:

Differentiating, $R^{'} (t) = 2 t - 6$ . Set equal to $0$ :

$2 t - 6 = 0 t = 3$

3. Identify extrema with the EVT:

Test the candidates (critical points and endpoints) into the original function $R (t)$ .

$R (0) = 0^{2} - 6 (0) + 10 = 4$
$R (3) = 3^{2} - 6 (3) + 10 = 1$
$R (4) = 4^{2} - 6 (4) + 10 = 2$

Therefore absolute minimum value of the rate, $1$ barrel per hour, over the specified time interval occurs at $t = 3$ hours.

Example 3: Motion

A particle moves along the $x$ -axis so that its velocity at time $t$ is given by $v (t) = - t^{3} + 6 t^{2} + 15 t$ for $t > 0$ . At what time $t$ does the particle achieve its maximum acceleration, and what is the maximum acceleration?

Solution

(spoiler)

1. Identify the objective function + constraint:

The question asks for maximum acceleration, meaning we want to find the maximum of the acceleration function. Differentiating velocity,

$a (t) = - 3 t^{2} + 12 t + 15$

There is no constraint equation since the function is already in terms of a single variable, $t$ .

2. Find the critical points:

Differentiate acceleration:

$a^{'} (t) = - 6 t + 12$

Set it equal to $0$ :

$- 6 t + 12 = 0 t = 2$

3. Justify extrema:

Now, check that $t = 2$ yields a maximum using the 2nd derivative test. Differentiating again,

$a^{''} (t) = - 6$

Because $a^{''} (t) < 0$ for all $t$ , then $a (t)$ is concave down at the critical point of $t = 2$ , proving a relative maximum.

Since $t = 2$ is the only critical point on the domain $t > 0$ , this relative maximum is automatically the absolute maximum. The maximum acceleration occurs at $t = 2$ .

Lastly, to find the maximum acceleration, find $a (2)$ :

$a (2) = - 3 (2)^{2} + 12 (2) + 15 = 27$

AP tip:

Pay careful attention to the phrasing on optimization problems with motion!

Try to reword the question into a target statement such as: “We want to find the maximum of [blank] function.”

The following examples will be explained further in the quiz questions. As a translation guide:

“At what time $t$ is the particle furthest to the right/left?”
- Optimize position $x (t)$ .
“Find the maximum speed.”
- Maximize $∣ v (t) ∣$ , meaning the largest magnitude regardless of sign.
“At what time $t$ is the particle’s velocity decreasing (or increasing) the fastest?”
- Optimize acceleration $a (t)$ .

Optimization strategy (4 steps)

Step 1: Identify objective function (quantity to maximize/minimize)
Step 2: Identify constraint equation(s)
Steps 3–4: Substitute constraint into objective, then find critical points via $f^{'} (x) = 0$ and classify with 2nd derivative test

Distinguishing optimization from related rates

Optimization keywords: maximum, minimum, largest, least
Related rates keywords: increasing/decreasing at a rate, changing with respect to time
Time-based optimization still possible if asked for when a function is maximized/minimized (not a rate)

Closed interval optimization (EVT)

Use EVT when problem asks for an absolute extremum on a closed interval $[a, b]$
Candidates: all critical points + both endpoints
Evaluate objective function at each candidate; compare to identify absolute max/min

Motion optimization

Match the objective function to what’s being optimized (e.g., max acceleration → differentiate $v (t)$ to get $a (t)$ , then optimize $a (t)$ )
If only one critical point exists on the domain, a relative extremum is automatically the absolute extremum
Always substitute back into the correct function to find the actual max/min value

Optimization

What you’ll learn

How to solve optimization problems with a 4-step strategy
How to apply the EVT to optimization problems
How to optimize motion functions

A common real-world application of derivative tests is optimization, which means finding the maximum or minimum value of a quantity while meeting certain constraints.

Example 1: General strategy

The following example will be used to show the step-by-step process:

Let $y = 12 - x^{2}$ for $x > 0$ .

Find the value of $x$ that maximizes the product of $x$ and $y$ .

Solution

Step 1: Identify the objective function + constraint equation

The objective function is the quantity to maximize or minimize. We want to maximize the product, $P$ :

$P = x y$

The constraint equation is the rule that limits your variables. Here, the constraint is given as a function of $x$ :

$y = 12 - x^{2}$

Step 2: Rewrite the objective function in terms of a single variable.

Substitute the constraint into the objective function.

$P (x) = x (12 - x^{2}) = 12 x - x^{3}$

Step 3: Find critical points:

To find where the product is maximized, find the critical points by taking the derivative and setting it equal to zero ( $P^{'} (x) = 0$ ). Differentiating,

$P^{'} (x) = 12 - 3 x^{2}$

Then solving for the critical points:

$12 - 3 x^{2} 124 x = 0 = 3 x^{2} = x^{2} = \pm 2$

Since the problem states $x > 0$ , our only critical point is $x = 2$ .

Step 4: Justify extrema of the objective function

Use either the 1st or 2nd derivative test to classify the critical point as extrema. Here, we’ll use the 2nd derivative to check that $x = 2$ yields a maximum. Differentiating again,

$P^{''} (x) = - 6 x$

Evaluating,

$P^{''} (2) = - 6 (2) = - 12$

Because $P^{''} (2) < 0$ , the function is concave down at this point, which proves that $x = 2$ produces a relative maximum.

Sidenote

Not related rates

Look for keywords that distinguish the two, such as

Optimization: Maximum, minimum, largest, or least
Related rates: Increasing at a rate, decreasing, or changing with respect to time

Example 2: Closed interval optimization

The rate at which a processing plant refines oil is modeled by the function $R (t) = t^{2} - 6 t + 10$ barrels per hour, where $t$ is measured in hours for $0 \leq t \leq 4$ . At what time $t$ is the refinery processing oil at the absolute slowest rate?

Solution

(spoiler)

The keyword “absolute” signals use of the EVT, while “slowest” frames it as an optimization problem to find the minimum of the function.

1. Identify the objective function + constraint:

We want to minimize the refining rate, $R (t)$ .

There is no constraint equation since the function is already in terms of a single variable, $t$ .

2. Find the critical points:

Differentiating, $R^{'} (t) = 2 t - 6$ . Set equal to $0$ :

$2 t - 6 = 0 t = 3$

3. Identify extrema with the EVT:

Test the candidates (critical points and endpoints) into the original function $R (t)$ .

$R (0) = 0^{2} - 6 (0) + 10 = 4$
$R (3) = 3^{2} - 6 (3) + 10 = 1$
$R (4) = 4^{2} - 6 (4) + 10 = 2$

Therefore absolute minimum value of the rate, $1$ barrel per hour, over the specified time interval occurs at $t = 3$ hours.

Example 3: Motion

A particle moves along the $x$ -axis so that its velocity at time $t$ is given by $v (t) = - t^{3} + 6 t^{2} + 15 t$ for $t > 0$ . At what time $t$ does the particle achieve its maximum acceleration, and what is the maximum acceleration?

Solution

(spoiler)

1. Identify the objective function + constraint:

The question asks for maximum acceleration, meaning we want to find the maximum of the acceleration function. Differentiating velocity,

$a (t) = - 3 t^{2} + 12 t + 15$

There is no constraint equation since the function is already in terms of a single variable, $t$ .

2. Find the critical points:

Differentiate acceleration:

$a^{'} (t) = - 6 t + 12$

Set it equal to $0$ :

$- 6 t + 12 = 0 t = 2$

3. Justify extrema:

Now, check that $t = 2$ yields a maximum using the 2nd derivative test. Differentiating again,

$a^{''} (t) = - 6$

Because $a^{''} (t) < 0$ for all $t$ , then $a (t)$ is concave down at the critical point of $t = 2$ , proving a relative maximum.

Since $t = 2$ is the only critical point on the domain $t > 0$ , this relative maximum is automatically the absolute maximum. The maximum acceleration occurs at $t = 2$ .

Lastly, to find the maximum acceleration, find $a (2)$ :

$a (2) = - 3 (2)^{2} + 12 (2) + 15 = 27$

AP tip:

Pay careful attention to the phrasing on optimization problems with motion!

Try to reword the question into a target statement such as: “We want to find the maximum of [blank] function.”

The following examples will be explained further in the quiz questions. As a translation guide:

“At what time $t$ is the particle furthest to the right/left?”
- Optimize position $x (t)$ .
“Find the maximum speed.”
- Maximize $∣ v (t) ∣$ , meaning the largest magnitude regardless of sign.
“At what time $t$ is the particle’s velocity decreasing (or increasing) the fastest?”
- Optimize acceleration $a (t)$ .

Achievable AP Calculus AB

5. Analytical uses

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.

Optimization

5 min read

Font

Discuss

Feedback

What you’ll learn

How to solve optimization problems with a 4-step strategy
How to apply the EVT to optimization problems
How to optimize motion functions

A common real-world application of derivative tests is optimization, which means finding the maximum or minimum value of a quantity while meeting certain constraints.

Example 1: General strategy

The following example will be used to show the step-by-step process:

Let $y = 12 - x^{2}$ for $x > 0$ .

Find the value of $x$ that maximizes the product of $x$ and $y$ .

Solution

Step 1: Identify the objective function + constraint equation

The objective function is the quantity to maximize or minimize. We want to maximize the product, $P$ :

$P = x y$

The constraint equation is the rule that limits your variables. Here, the constraint is given as a function of $x$ :

$y = 12 - x^{2}$

Step 2: Rewrite the objective function in terms of a single variable.

Substitute the constraint into the objective function.

$P (x) = x (12 - x^{2}) = 12 x - x^{3}$

Step 3: Find critical points:

To find where the product is maximized, find the critical points by taking the derivative and setting it equal to zero ( $P^{'} (x) = 0$ ). Differentiating,

$P^{'} (x) = 12 - 3 x^{2}$

Then solving for the critical points:

$12 - 3 x^{2} 124 x = 0 = 3 x^{2} = x^{2} = \pm 2$

Since the problem states $x > 0$ , our only critical point is $x = 2$ .

Step 4: Justify extrema of the objective function

Use either the 1st or 2nd derivative test to classify the critical point as extrema. Here, we’ll use the 2nd derivative to check that $x = 2$ yields a maximum. Differentiating again,

$P^{''} (x) = - 6 x$

Evaluating,

$P^{''} (2) = - 6 (2) = - 12$

Because $P^{''} (2) < 0$ , the function is concave down at this point, which proves that $x = 2$ produces a relative maximum.

Sidenote

Not related rates

Look for keywords that distinguish the two, such as

Optimization: Maximum, minimum, largest, or least
Related rates: Increasing at a rate, decreasing, or changing with respect to time

Example 2: Closed interval optimization

The rate at which a processing plant refines oil is modeled by the function $R (t) = t^{2} - 6 t + 10$ barrels per hour, where $t$ is measured in hours for $0 \leq t \leq 4$ . At what time $t$ is the refinery processing oil at the absolute slowest rate?

Solution

(spoiler)

The keyword “absolute” signals use of the EVT, while “slowest” frames it as an optimization problem to find the minimum of the function.

1. Identify the objective function + constraint:

We want to minimize the refining rate, $R (t)$ .

There is no constraint equation since the function is already in terms of a single variable, $t$ .

2. Find the critical points:

Differentiating, $R^{'} (t) = 2 t - 6$ . Set equal to $0$ :

$2 t - 6 = 0 t = 3$

3. Identify extrema with the EVT:

Test the candidates (critical points and endpoints) into the original function $R (t)$ .

$R (0) = 0^{2} - 6 (0) + 10 = 4$
$R (3) = 3^{2} - 6 (3) + 10 = 1$
$R (4) = 4^{2} - 6 (4) + 10 = 2$

Therefore absolute minimum value of the rate, $1$ barrel per hour, over the specified time interval occurs at $t = 3$ hours.

Example 3: Motion

A particle moves along the $x$ -axis so that its velocity at time $t$ is given by $v (t) = - t^{3} + 6 t^{2} + 15 t$ for $t > 0$ . At what time $t$ does the particle achieve its maximum acceleration, and what is the maximum acceleration?

Solution

(spoiler)

1. Identify the objective function + constraint:

The question asks for maximum acceleration, meaning we want to find the maximum of the acceleration function. Differentiating velocity,

$a (t) = - 3 t^{2} + 12 t + 15$

There is no constraint equation since the function is already in terms of a single variable, $t$ .

2. Find the critical points:

Differentiate acceleration:

$a^{'} (t) = - 6 t + 12$

Set it equal to $0$ :

$- 6 t + 12 = 0 t = 2$

3. Justify extrema:

Now, check that $t = 2$ yields a maximum using the 2nd derivative test. Differentiating again,

$a^{''} (t) = - 6$

Because $a^{''} (t) < 0$ for all $t$ , then $a (t)$ is concave down at the critical point of $t = 2$ , proving a relative maximum.

Since $t = 2$ is the only critical point on the domain $t > 0$ , this relative maximum is automatically the absolute maximum. The maximum acceleration occurs at $t = 2$ .

Lastly, to find the maximum acceleration, find $a (2)$ :

$a (2) = - 3 (2)^{2} + 12 (2) + 15 = 27$

AP tip:

Pay careful attention to the phrasing on optimization problems with motion!

Try to reword the question into a target statement such as: “We want to find the maximum of [blank] function.”

The following examples will be explained further in the quiz questions. As a translation guide:

“At what time $t$ is the particle furthest to the right/left?”
- Optimize position $x (t)$ .
“Find the maximum speed.”
- Maximize $∣ v (t) ∣$ , meaning the largest magnitude regardless of sign.
“At what time $t$ is the particle’s velocity decreasing (or increasing) the fastest?”
- Optimize acceleration $a (t)$ .

Optimization strategy (4 steps)

Step 1: Identify objective function (quantity to maximize/minimize)
Step 2: Identify constraint equation(s)
Steps 3–4: Substitute constraint into objective, then find critical points via $f^{'} (x) = 0$ and classify with 2nd derivative test

Distinguishing optimization from related rates

Optimization keywords: maximum, minimum, largest, least
Related rates keywords: increasing/decreasing at a rate, changing with respect to time
Time-based optimization still possible if asked for when a function is maximized/minimized (not a rate)

Closed interval optimization (EVT)

Use EVT when problem asks for an absolute extremum on a closed interval $[a, b]$
Candidates: all critical points + both endpoints
Evaluate objective function at each candidate; compare to identify absolute max/min

Motion optimization

Match the objective function to what’s being optimized (e.g., max acceleration → differentiate $v (t)$ to get $a (t)$ , then optimize $a (t)$ )
If only one critical point exists on the domain, a relative extremum is automatically the absolute extremum
Always substitute back into the correct function to find the actual max/min value

Optimization

What you’ll learn

How to solve optimization problems with a 4-step strategy
How to apply the EVT to optimization problems
How to optimize motion functions

A common real-world application of derivative tests is optimization, which means finding the maximum or minimum value of a quantity while meeting certain constraints.

Example 1: General strategy

The following example will be used to show the step-by-step process:

Let $y = 12 - x^{2}$ for $x > 0$ .

Find the value of $x$ that maximizes the product of $x$ and $y$ .

Solution

Step 1: Identify the objective function + constraint equation

The objective function is the quantity to maximize or minimize. We want to maximize the product, $P$ :

$P = x y$

The constraint equation is the rule that limits your variables. Here, the constraint is given as a function of $x$ :

$y = 12 - x^{2}$

Step 2: Rewrite the objective function in terms of a single variable.

Substitute the constraint into the objective function.

$P (x) = x (12 - x^{2}) = 12 x - x^{3}$

Step 3: Find critical points:

To find where the product is maximized, find the critical points by taking the derivative and setting it equal to zero ( $P^{'} (x) = 0$ ). Differentiating,

$P^{'} (x) = 12 - 3 x^{2}$

Then solving for the critical points:

$12 - 3 x^{2} 124 x = 0 = 3 x^{2} = x^{2} = \pm 2$

Since the problem states $x > 0$ , our only critical point is $x = 2$ .

Step 4: Justify extrema of the objective function

Use either the 1st or 2nd derivative test to classify the critical point as extrema. Here, we’ll use the 2nd derivative to check that $x = 2$ yields a maximum. Differentiating again,

$P^{''} (x) = - 6 x$

Evaluating,

$P^{''} (2) = - 6 (2) = - 12$

Because $P^{''} (2) < 0$ , the function is concave down at this point, which proves that $x = 2$ produces a relative maximum.

Sidenote

Not related rates

Look for keywords that distinguish the two, such as

Optimization: Maximum, minimum, largest, or least
Related rates: Increasing at a rate, decreasing, or changing with respect to time

Example 2: Closed interval optimization

The rate at which a processing plant refines oil is modeled by the function $R (t) = t^{2} - 6 t + 10$ barrels per hour, where $t$ is measured in hours for $0 \leq t \leq 4$ . At what time $t$ is the refinery processing oil at the absolute slowest rate?

Solution

(spoiler)

The keyword “absolute” signals use of the EVT, while “slowest” frames it as an optimization problem to find the minimum of the function.

1. Identify the objective function + constraint:

We want to minimize the refining rate, $R (t)$ .

There is no constraint equation since the function is already in terms of a single variable, $t$ .

2. Find the critical points:

Differentiating, $R^{'} (t) = 2 t - 6$ . Set equal to $0$ :

$2 t - 6 = 0 t = 3$

3. Identify extrema with the EVT:

Test the candidates (critical points and endpoints) into the original function $R (t)$ .

$R (0) = 0^{2} - 6 (0) + 10 = 4$
$R (3) = 3^{2} - 6 (3) + 10 = 1$
$R (4) = 4^{2} - 6 (4) + 10 = 2$

Therefore absolute minimum value of the rate, $1$ barrel per hour, over the specified time interval occurs at $t = 3$ hours.

Example 3: Motion

A particle moves along the $x$ -axis so that its velocity at time $t$ is given by $v (t) = - t^{3} + 6 t^{2} + 15 t$ for $t > 0$ . At what time $t$ does the particle achieve its maximum acceleration, and what is the maximum acceleration?

Solution

(spoiler)

1. Identify the objective function + constraint:

The question asks for maximum acceleration, meaning we want to find the maximum of the acceleration function. Differentiating velocity,

$a (t) = - 3 t^{2} + 12 t + 15$

There is no constraint equation since the function is already in terms of a single variable, $t$ .

2. Find the critical points:

Differentiate acceleration:

$a^{'} (t) = - 6 t + 12$

Set it equal to $0$ :

$- 6 t + 12 = 0 t = 2$

3. Justify extrema:

Now, check that $t = 2$ yields a maximum using the 2nd derivative test. Differentiating again,

$a^{''} (t) = - 6$

Because $a^{''} (t) < 0$ for all $t$ , then $a (t)$ is concave down at the critical point of $t = 2$ , proving a relative maximum.

Since $t = 2$ is the only critical point on the domain $t > 0$ , this relative maximum is automatically the absolute maximum. The maximum acceleration occurs at $t = 2$ .

Lastly, to find the maximum acceleration, find $a (2)$ :

$a (2) = - 3 (2)^{2} + 12 (2) + 15 = 27$

AP tip:

Pay careful attention to the phrasing on optimization problems with motion!

Try to reword the question into a target statement such as: “We want to find the maximum of [blank] function.”

The following examples will be explained further in the quiz questions. As a translation guide:

“At what time $t$ is the particle furthest to the right/left?”
- Optimize position $x (t)$ .
“Find the maximum speed.”
- Maximize $∣ v (t) ∣$ , meaning the largest magnitude regardless of sign.
“At what time $t$ is the particle’s velocity decreasing (or increasing) the fastest?”
- Optimize acceleration $a (t)$ .