Solving Optimization Problems - AP Calculus BC Study Guide

The Core Idea: Solving Optimization Problems

Optimization is the process of finding the absolute maximum or absolute minimum value of a function, which often represents a real-world quantity such as area, volume, cost, or distance. The fundamental tool for solving these problems is the derivative. By finding where the derivative of a function is zero or undefined, we can identify critical points that are candidates for maximum or minimum values.

The core task in an optimization problem is to translate a verbal description or a given scenario into a mathematical function of a single variable. This often involves identifying a primary equation for the quantity to be optimized and a secondary (or constraint) equation that relates the variables. Once this function is established, along with its practical domain, calculus techniques are used to locate and justify the absolute extremum.

Key Theorems & Tests

The process of solving optimization problems relies on the following fundamental theorems and tests.

The Extreme Value Theorem (EVT)

If a function $f$ is continuous on a closed interval $[a,b]$, then $f$ is guaranteed to have both an absolute maximum and an absolute minimum value on that interval. These absolute extrema must occur either at a critical point of $f$ within the interval $(a,b)$ or at one of the endpoints, $a$ or $b$.

The First Derivative Test

Let $c$ be a critical point of a function $f$.

• 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 $c$ is the only critical point on an interval, this test can be used to justify an absolute extremum.

The Second Derivative Test

Let $c$ be a critical point of a function $f$ such that $f^{\prime }(c)=0$.

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

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

• If $f^{\prime \prime }(c)=0$, the test is inconclusive.

Similar to the First Derivative Test, this can be used to justify an absolute extremum if $c$ is the only critical point on an interval.

Understanding the Problem-Solving Process

The most challenging part of optimization is often setting up the problem correctly. The process involves several distinct steps before any calculus is performed. First, one must carefully read the problem to identify the quantity to be maximized or minimized (e.g., "maximize area," "minimize cost"). This quantity will be represented by a primary function.

Next, identify all variables in the problem. In many cases, the primary function will initially depend on more than one variable. To proceed, you must find a constraint or relationship between the variables given in the problem statement (e.g., "the perimeter is 100 feet," "the volume is 50 cm^3"). This constraint provides a secondary equation. The secondary equation is used to solve for one variable in terms of another, allowing you to substitute it into the primary equation. The result is the function to be optimized, now expressed in terms of a single variable. Finally, you must determine the domain of this function—the set of all possible input values that are physically realistic for the problem. This domain can be a closed or open interval and is critical for choosing the correct justification method.

Core Concepts & Rules

• The Goal of Optimization: To find the absolute maximum or minimum value of a function on a given interval, which may be open or closed.

• Constructing the Function: The first step is to create a function for the quantity to be optimized. This may involve using a given formula or constructing one from a verbal description.

• Single-Variable Analysis: If the function depends on multiple variables, use a given constraint to rewrite the function in terms of a single variable.

• Determining the Domain: Identify the practical domain for the single-variable function based on the physical constraints of the problem (e.g., lengths cannot be negative).

• Optimization on a Closed Interval $[a,b]$: For a continuous function on a closed interval, the Extreme Value Theorem applies. The absolute extremum is found by testing the function's values at all critical points within the interval and at the endpoints $a$ and $b$. The largest value is the absolute maximum, and the smallest is the absolute minimum.

• Optimization on an Open Interval: If the domain is not a closed interval, you must use the First or Second Derivative Test to justify that a critical point corresponds to an absolute extremum. This typically involves finding a single critical point in the interval and showing that it yields the desired maximum or minimum.

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

A farmer has 2400 ft of fencing and wants to fence off a rectangular field that borders a straight river. He needs no fence along the river. What are the dimensions of the field that has the largest area?

Step 1: Identify the primary equation (to be maximized).

We want to maximize the area $A$ of the rectangle. Let $x$ be the length of the side perpendicular to the river and $y$ be the length of the side parallel to the river.

The area is $A=xy$.

Step 2: Identify the secondary equation (the constraint).

The amount of fencing is limited to 2400 ft. The fence consists of two sides of length $x$ and one side of length $y$.

The constraint is $2x+y=2400$.

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

From the constraint, solve for $y$: $y=2400-2x$.

Substitute this into the area equation:

$A(x)=x(2400-2x)=2400x-2x^2$.

Step 4: Determine the domain of the function.

Since $x$ represents a physical length, $x\ge 0$.

Since $y$ must also be a non-negative length, $y=2400-2x\ge 0$, which implies $2400\ge 2x$, or $x\le 1200$.

The domain is a closed interval: $[0,1200]$.

Step 5: Find the derivative and critical points.

$A^{\prime }(x)=\frac{d}{dx}(2400x-2x^2)=2400-4x$.

Set $A^{\prime }(x)=0$ to find critical points:

$2400-4x=0⟹4x=2400⟹x=600$.

The only critical point in the interval $(0,1200)$ is $x=600$.

Step 6: Justify the absolute maximum using the Extreme Value Theorem.

Since $A(x)$ is a continuous function on the closed interval $[0,1200]$, we test the critical point and the endpoints.

• Endpoint: $A(0)=2400(0)-2(0{)}^2=0$

• Critical Point: $A(600)=2400(600)-2(600{)}^2=1,440,000-720,000=720,000$

• Endpoint: $A(1200)=2400(1200)-2(1200{)}^2=2,880,000-2,880,000=0$

Step 7: State the final answer.

The maximum area is 720,000 sq ft. This occurs when $x=600$ ft.

The corresponding $y$ dimension is $y=2400-2(600)=2400-1200=1200$ ft.

The dimensions of the field are 600 ft by 1200 ft.

Step-by-Step Example 2: Optimization on an Open Interval

We want to construct a cylindrical can with a fixed volume of $1000{\text{ cm}}^3$. What are the dimensions (radius and height) that will minimize the surface area of the can?

Step 1: Identify the primary equation (to be minimized).

We want to minimize the surface area $S$ of a cylinder. The area consists of the top and bottom circles ($2\pi r^2$) and the side ($2\pi rh$).

$S=2\pi r^2+2\pi rh$.

Step 2: Identify the secondary equation (the constraint).

The volume $V$ is fixed at $1000{\text{ cm}}^3$. The formula for the volume of a cylinder is $V=\pi r^{2}h$.

The constraint is $\pi r^{2}h=1000$.

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

From the constraint, solve for $h$: $h=\frac{1000}{\pi r^2}$.

Substitute this into the surface area equation:

$S(r)=2\pi r^2+2\pi r\left(\frac{1000}{\pi r^2}\right)=2\pi r^2+\frac{2000}{r}$.

Step 4: Determine the domain of the function.

The radius $r$ must be positive, so $r>0$. The height $h$ must also be positive, which is guaranteed if $r>0$.

The domain is an open interval: $(0,\infty )$.

Step 5: Find the derivative and critical points.

Rewrite $S(r)=2\pi r^2+2000r^{-1}$.

$S^{\prime }(r)=4\pi r-2000r^{-2}=4\pi r-\frac{2000}{r^2}$.

Set $S^{\prime }(r)=0$:

$4\pi r=\frac{2000}{r^2}⟹4\pi r^3=2000⟹r^3=\frac{2000}{4\pi }=\frac{500}{\pi }$.

The only critical point in the domain is $r=\sqrt[3]{\frac{500}{\pi }}$.

Step 6: Justify the absolute minimum using the First or Second Derivative Test.

Since the domain is an open interval, we cannot use the EVT. We will use the Second Derivative Test.

$S^{\prime \prime }(r)=\frac{d}{dr}(4\pi r-2000r^{-2})=4\pi +4000r^{-3}=4\pi +\frac{4000}{r^3}$.

For any $r>0$, $S^{\prime \prime }(r)$ is positive ($4\pi >0$ and $\frac{4000}{r^3}>0$).

Therefore, at the critical point $r=\sqrt[3]{\frac{500}{\pi }}$, $S^{\prime \prime }(r)>0$.

By the Second Derivative Test, this critical point corresponds to a local minimum. Since it is the only critical point on the open interval $(0,\infty )$, it must be the location of the absolute minimum.

Step 7: State the final answer.

The surface area is minimized when $r=\sqrt[3]{\frac{500}{\pi }}$ cm.

The corresponding height is $h=\frac{1000}{\pi r^2}=\frac{1000}{\pi {\left(\frac{500}{\pi }\right)}^{2/3}}$.

This simplifies to $h=2\sqrt[3]{\frac{500}{\pi }}=2r$.

The dimensions that minimize the surface area are $r=\sqrt[3]{\frac{500}{\pi }}$ cm and $h=2\sqrt[3]{\frac{500}{\pi }}$ cm.

Using Your Calculator

Optimization problems are primarily solved through analytical methods (using derivatives and justification tests). A calculator should be seen as a tool to aid in computation or to verify a result, not as the primary method for solving.

1. Finding Critical Points: If setting the derivative equal to zero results in an equation that is difficult to solve algebraically, you can use your calculator.

   • Graph the derivative function, $y_1=f^{\prime }(x)$.

   • Use the CALC -> $zero$ feature to find the x-intercepts, which correspond to the critical points.

2. Verifying Extrema: You can verify your final answer by graphing the original function you constructed, $y_1=f(x)$.

   • Set the window to match the domain of your problem.

   • Use the CALC -> $minimum$ or $CALC$ -> $maximum$ feature to find the coordinates of the extremum. This can confirm the location ($x$) and value ($f(x)$) you found analytically.

3. Checking with Derivative Tests: The calculator can help apply the First Derivative Test.

   • After finding a critical point $c$, you can evaluate the derivative $f^{\prime }(x)$ at test points on either side of $c$. For example, use the calculator to compute $f^{\prime }(c-0.1)$ and $f^{\prime }(c+0.1)$ to check for a sign change.

Important: On the AP Exam, you must show the analytical work. A calculator answer without the underlying calculus (the derivative, the setup, and the justification) will not receive full credit.

AP Exam Quick Hit

Common Question Types

• Geometric Optimization: Find the maximum area of a rectangle inscribed between the x-axis and a given curve, such as $f(x)=9-x^2$. This requires setting up an area function $A(x)=(\text{width})(\text{height})=(2x)(9-x^2)$ on a closed domain like $[0,3]$.

• Distance Minimization: Find the point on the graph of a function, like $y=x^2$, that is closest to a given point, such as $(0,5)$. This involves minimizing the distance formula, or more simply, minimizing the square of the distance: $D^2=(x-0{)}^2+(x^2-5{)}^2$.

• Cost/Resource Minimization: A problem involving minimizing cost or materials, such as finding the dimensions of a box with a fixed volume and a square base that minimizes material cost, where the top and bottom material costs differ from the side material.

Common Mistakes

• Forgetting Justification: The most common error is finding the correct critical point but failing to provide a mathematical justification (using the EVT, First Derivative Test, or Second Derivative Test) that it yields an absolute maximum or minimum.

• Answering the Wrong Question: After finding the value of the variable that optimizes the function (e.g., $x=5$), students might provide this as the answer instead of the actual maximum or minimum value (e.g., the maximum area $A(5)$). Always re-read the prompt.

• Ignoring Endpoints: When working on a closed interval, students often find the critical point and assume it is the answer without testing the function's value at the endpoints. The absolute extremum could occur at an endpoint.

• Incorrect Setup: Errors made in translating the word problem into the primary and secondary equations are very common. Drawing a diagram can help prevent these setup mistakes.

• Assuming a Relative Extremum is Absolute: On an open interval, finding a local minimum using a derivative test is not sufficient. You must state that because it is the only critical point on the interval, it is the absolute minimum.