Introduction to Optimization Problems - AP Calculus BC Study Guide

The Core Idea: Introduction to Optimization Problems

Optimization is the process of finding the absolute maximum or absolute minimum value of a function, which represents a real-world quantity we wish to maximize (like profit or volume) or minimize (like cost or surface area). The fundamental tool for optimization is the derivative. By analyzing the derivative of a function, we can identify critical points—locations where the function might attain a maximum or minimum value.

The core of optimization involves a systematic process: modeling a situation with a function, finding the critical points of that function using its derivative, and then applying rigorous analytical tests to determine which of these points (or the endpoints of the interval) yields the absolute maximum or minimum value sought. The choice of method depends on whether the function is considered over a closed interval, where the Candidates Test is effective, or an open interval, where derivative tests are used to justify an absolute extremum.

Key Definitions and Tests

Critical Point

A critical point of a function $f(x)$ is a point $x=c$ in the domain of $f$ where either $f^{\prime }(c)=0$ or $f^{\prime }(c)$ is undefined. These are the only points where a function can have a local (relative) maximum or minimum. The derivative is the tool used to locate these potential points of interest.

The Candidates Test

The Candidates Test is a method for finding the absolute maximum and minimum values of a continuous function on a closed interval$[a,b]$. The process is as follows:

1. Find all critical points of the function in the open interval $(a,b)$.

2. Create a list of "candidates" consisting of the endpoints $a$ and $b$, and all critical points found in step 1.

3. Evaluate the function $f(x)$ at every candidate value from step 2.

4. The largest function value is the absolute maximum, and the smallest function value is the absolute minimum on the interval $[a,b]$.

The First Derivative Test

The First Derivative Test is used to determine whether a critical point is a local maximum, a local minimum, or neither. Let $c$ be a critical point of $f(x)$:

• If $f^{\prime }(x)$ changes from positive to negative at $x=c$, then $f$ has a local maximum at $c$.

• If $f^{\prime }(x)$ changes from negative to positive at $x=c$, then $f$ has a local minimum at $c$.

• If $f^{\prime }(x)$ does not change sign at $x=c$, then $f$ has neither a local maximum nor a local minimum at $c$.

The Second Derivative Test

The Second Derivative Test is another method to classify a critical point, specifically one where $f^{\prime }(c)=0$. Let $c$ be such a critical point:

• If $f^{\prime \prime }(c)>0$, then $f$ has a local minimum at $c$. (The function is concave up).

• If $f^{\prime \prime }(c)<0$, then $f$ has a local maximum at $c$. (The function is concave down).

• If $f^{\prime \prime }(c)=0$, the test is inconclusive, and the First Derivative Test must be used.

Understanding Justification for Absolute Extrema

A key nuance in optimization is justifying that a found extremum is absolute, not just relative (local). The method of justification depends on the interval of interest.

For a continuous function on a closed interval $[a,b]$, the absolute extrema are guaranteed to exist by the Extreme Value Theorem. The Candidates Test is the definitive method. By evaluating the function at all critical points and at the endpoints, you are guaranteed to find the absolute maximum and minimum values.

For a continuous function on an open interval, the situation is different. There may not be an absolute maximum or minimum. However, a special and common case arises in optimization problems: if a function has only one critical point within an interval, and that critical point corresponds to a local maximum or minimum (as verified by the First or Second Derivative Test), then that local extremum is also the absolute extremum on that interval. This is a critical justification step for many word problems where the domain is naturally an open interval (e.g., lengths cannot be zero or a certain maximum value).

Core Concepts & Rules

• Locating Potential Extrema: The derivative, $f^{\prime }(x)$, is used to find the critical points of a function $f(x)$. These are the only locations where local extrema can occur.

• Absolute Extrema on a Closed Interval: To find the absolute maximum and minimum of a continuous function on $[a,b]$, always use the Candidates Test. This involves comparing the function's values at the endpoints ($f(a)$ and $f(b)$) and at all critical points inside the interval.

• Classifying Critical Points: The First Derivative Test (analyzing the sign of $f^{\prime }(x)$) and the Second Derivative Test (analyzing the sign of $f^{\prime \prime }(x)$ at the critical point) are used to classify a critical point as a local maximum or a local minimum.

• Justifying Absolute Extrema on an Open Interval: If a continuous function has exactly one critical point on an interval, and you can show that this point is a local maximum (or minimum) using a derivative test, then you can conclude it is the absolute maximum (or minimum) on that interval.

Step-by-Step Example 1: Optimization on a Closed Interval

Problem: Find the absolute maximum and minimum values of the function $f(x)=2x^3-3x^2-12x+1$ on the closed interval $[-2,3]$.

Step 1: Find the derivative.

The derivative is used to locate the critical points.

$$f^{\prime }(x)=6x^2-6x-12$$

Step 2: Find the critical points.

Set the derivative equal to zero and solve for $x$. The derivative is a polynomial and is never undefined.

$$6x^2-6x-12=0$$

$$6(x^2-x-2)=0$$

$$6(x-2)(x+1)=0$$

The critical points are $x=2$ and $x=-1$. Both of these points are within the interval $[-2,3]$.

Step 3: Apply the Candidates Test.

The candidates for the absolute extrema are the endpoints and the critical points within the interval.

• Endpoint: $x=-2$

• Critical Point: $x=-1$

• Critical Point: $x=2$

• Endpoint: $x=3$

Step 4: Evaluate the function at each candidate.

• $f(-2)=2(-2{)}^3-3(-2{)}^2-12(-2)+1=-16-12+24+1=-3$

• $f(-1)=2(-1{)}^3-3(-1{)}^2-12(-1)+1=-2-3+12+1=8$

• $f(2)=2(2{)}^3-3(2{)}^2-12(2)+1=16-12-24+1=-19$

• $f(3)=2(3{)}^3-3(3{)}^2-12(3)+1=54-27-36+1=-8$

Step 5: Conclude the absolute maximum and minimum.

By comparing the output values:

• The absolute maximum value is 8, which occurs at $x=-1$.

• The absolute minimum value is -19, which occurs at $x=2$.

Step-by-Step Example 2: Exam-Style Application

Problem: A rectangular garden is to be constructed using a rock wall as one side and wire fencing for the other three sides. If 100 feet of wire fencing is available, what is the largest possible area of the garden?

Step 1: Define variables and the primary equation.

Let $x$ be the length of the two sides perpendicular to the rock wall, and let $y$ be the length of the side parallel to the wall.

The quantity to be maximized is the area, $A$.

Primary Equation: $A=xy$

Step 2: Write the secondary (constraint) equation.

The constraint is the amount of fencing available.

Secondary Equation: $2x+y=100$

Step 3: Express the primary equation in terms of a single variable.

Solve the secondary equation for $y$: $y=100-2x$.

Substitute this into the area equation:

$A(x)=x(100-2x)=100x-2x^2$

Step 4: Determine the relevant domain.

Since $x$ represents a length, $x>0$. Also, the parallel side $y$ must be positive, so $100-2x>0$, which means $100>2x$, or $x<50$.

The domain is the open interval $(0,50)$.

Step 5: Find the derivative of the function to be optimized.

$A^{\prime }(x)=100-4x$

Step 6: Find the critical points.

Set the derivative to zero and solve.

$100-4x=0⟹4x=100⟹x=25$

This single critical point, $x=25$, is within our domain $(0,50)$.

Step 7: Justify that the critical point yields an absolute maximum.

We will use the Second Derivative Test.

$A^{\prime \prime }(x)=-4$

Since $A^{\prime \prime }(25)=-4<0$, the function $A(x)$ has a local maximum at $x=25$.

Because $x=25$ is the only critical point on the open interval $(0,50)$, this local maximum must also be the absolute maximum.

Step 8: Answer the question.

The question asks for the largest possible area. We found the $x$ value that maximizes the area is $x=25$ feet.

The corresponding area is:

$A(25)=100(25)-2(25{)}^2=2500-2(625)=2500-1250=1250$

The largest possible area of the garden is 1250 square feet.

Using Your Calculator

While the analytical justification for optimization must be shown on the exam, a graphing calculator is a powerful tool for verifying your results.

To solve the problem $f(x)=2x^3-3x^2-12x+1$ on $[-2,3]$:

1. Graph the Function:

   • Press Y= and enter Y_1 = 2X^3 - 3X^2 - 12X + 1.

   • Set the WINDOW to match the interval: $Xmin=-2$, $Xmax=3$. Use $ZoomFit$ (often ZOOM -> $0$) to adjust the Y-axis, or set it manually to see the graph clearly (e.g., $Ymin=-25$, $Ymax=10$).

   • The graph will visually confirm a maximum near $x=-1$ and a minimum near $x=2$.

2. Find Critical Points:

   • You can graph the derivative $Y_2=6X^2-6X-12$ (the derivative of $Y_1$).

   • Use the CALC$ menu ($2nd$ + TRACE`) and select $2:zero$.

   • Use the cursor to find the x-intercepts of $Y_2$. The calculator will confirm the zeros at $x=-1$ and $x=2$.

3. Verify Extrema:

   • Go back to the graph of $Y_1$.

   • Use the CALC$ menu and select $4:maximum$. Move the cursor to the left of the peak, press ENTER, move to the right, press ENTER, and press ENTER` again for a guess. The calculator will show a maximum at or near $(-1,8)$.

   • Use the $CALC$ menu and select $3:minimum$. Repeat the process for the valley. The calculator will show a minimum at or near $(2,-19)$.

The calculator is excellent for checking the values you found analytically through the Candidates Test, but it is not a substitute for showing the calculus steps.

AP Exam Quick Hit

Common Question Types

• Find Absolute Extrema on a Closed Interval: A straightforward question asking for the absolute maximum and minimum value of a given function $f(x)$ on a specified interval $[a,b]$. This is a direct application of the Candidates Test.

• Optimization Word Problem: A problem describing a physical situation (e.g., minimizing surface area of a can, maximizing the volume of a box, finding the closest point on a curve to a given point) that requires you to set up the function, determine the domain, find the single critical point, and justify that it yields an absolute extremum.

• Analysis of $f$ from the Graph of $f^{\prime }$: You are given the graph of the derivative, $f^{\prime }(x)$, on an interval $[a,b]$. You are asked to find the x-coordinate where $f(x)$ attains its absolute maximum. This requires identifying critical points (where $f^{\prime }(x)=0$ or changes sign) and endpoints as candidates, and then using areas under the $f^{\prime }(x)$ graph to compare the change in $f(x)$ between the candidates.

Common Mistakes

• Forgetting to Test Endpoints: When applying the Candidates Test for a closed interval, students find the critical points but forget to evaluate the function at the endpoints, $f(a)$ and $f(b)$. The absolute extremum can, and often does, occur at an endpoint.

• Incomplete Justification: Stating that a critical point is a maximum or minimum without providing a valid calculus-based reason. You must explicitly use the First or Second Derivative Test and, for open intervals, state why a local extremum is also an absolute extremum (i.e., it's the only critical point in the interval).

• Answering the Wrong Question: In a word problem, correctly finding the value of $x$ that optimizes a quantity but failing to use that $x$ to find the requested maximum area, minimum cost, or other dimension. Always re-read the prompt before writing your final answer.

• Confusing $f$, $f^{\prime }$, and $f^{\prime \prime }$: Using the sign of $f(x)$ instead of $f^{\prime }(x)$ to find where a function is increasing, or stating that $f^{\prime \prime }(x)=0$ implies a point of inflection when it is only a possibility.

• Assuming a Critical Point is an Extremum: Finding where $f^{\prime }(x)=0$ and immediately assuming it must be the maximum or minimum without performing the First or Second Derivative Test to confirm.