Solving Optimization Problems - AP Calculus AB Study Guide

The Core Idea: Solving Optimization Problems

Optimization is the process of finding the maximum or minimum value of a function under a specific set of conditions or constraints. In calculus, we leverage the derivative to solve these applied problems. The core idea is that the maximum or minimum value of a differentiable function must occur at a point where its derivative is zero or undefined (a critical point), or at an endpoint of its domain.

By translating a real-world problem—such as maximizing the volume of a container, minimizing the cost of materials, or finding the shortest distance between a point and a curve—into a mathematical function, we can use the tools of differential calculus to find the optimal solution. This involves identifying the quantity to be optimized, expressing it as a function of a single variable, and then analyzing that function's derivative to locate and justify the required maximum or minimum value.

Key Concepts: The Optimization Process

The Essential Knowledge for this topic outlines a structured, multi-step process for solving optimization problems. There are no new formulas to memorize; rather, the key is to master this analytical procedure.

The process is as follows:

1. Understand the Problem: Read the problem carefully to identify the specific quantity that needs to be maximized or minimized. Identify all given information and any constraints.

2. Write a Primary Equation: This is the equation for the quantity you wish to optimize. For example, if you are maximizing area, this is your area formula ($A=l\cdot w$). This equation may initially involve more than one independent variable.

3. Write a Secondary Equation (Constraint): If the primary equation has more than one variable, you must find a relationship between those variables. This relationship, given by the problem's constraints, is the secondary equation. For example, a fixed amount of fencing would be a perimeter constraint ($P=2l+2w$).

4. Express the Primary Equation in Terms of a Single Variable: Use the secondary equation to solve for one variable and substitute it into the primary equation. This creates a function of a single variable that you can analyze using calculus.

5. Determine a Relevant Domain: Based on the physical constraints of the problem, determine the feasible domain for your single-variable function. For instance, lengths, areas, and volumes cannot be negative. This domain may be a closed or open interval.

6. Use Calculus to Find the Optimal Value:

   • Find the derivative of your single-variable function.

   • Find the critical points by setting the derivative equal to zero and finding where it is undefined.

   • Evaluate the function at the critical points and at the endpoints of the domain (if it is a closed interval).

   • Justify that your result is a maximum or minimum using the First Derivative Test or the Second Derivative Test.

Understanding the Domain and Justification

A critical nuance in solving optimization problems is the careful consideration of the domain and the formal justification of your answer.

The Domain: The domain represents the set of all possible input values (e.g., length, radius, time) that are physically realistic for the problem. For example, if $x$ represents the side length of a box, the domain must be $x>0$. If you are cutting squares of side length $x$ from the corners of a 10-inch by 12-inch piece of cardboard, the amount you cut, $2x$, cannot exceed the shorter side, so $2x<10$, which means $0<x<5$. Establishing this domain is crucial because the absolute maximum or minimum might occur at an endpoint of a closed interval domain, not just at a critical point.

Justification: Finding a critical point is not enough. The AP exam requires you to prove that this critical point yields the desired maximum or minimum. There are two primary methods for this justification:

1. The First Derivative Test: Analyze the sign of the derivative ($f^{\prime }(x)$) on either side of the critical point $c$.

   • For a relative maximum, $f^{\prime }(x)$ must change from positive to negative at $x=c$.

   • For a relative minimum, $f^{\prime }(x)$ must change from negative to positive at $x=c$.

2. The Second Derivative Test: Find the second derivative ($f^{\prime \prime }(x)$) and evaluate its sign at the critical point $c$ (where $f^{\prime }(c)=0$).

   • If $f^{\prime \prime }(c)<0$, the function is concave down, and $f(c)$ is a relative maximum.

   • If $f^{\prime \prime }(c)>0$, the function is concave up, and $f(c)$ is a relative minimum.

If the domain is a closed interval $[a,b]$, you must also use the Candidates Test. This involves evaluating the function at all critical points within the interval and at the endpoints $a$ and $b$. The largest of these values is the absolute maximum, and the smallest is the absolute minimum.

Core Concepts & Rules

• The derivative is the fundamental tool used to solve optimization problems by locating potential maximum and minimum values.

• The first step in any optimization problem is to define a primary equation, which is a function representing the quantity to be maximized or minimized.

• If the primary equation contains more than one independent variable, a secondary equation must be established based on the constraints of the problem.

• The secondary equation is used to rewrite the primary equation as a function of a single variable.

• A practical and relevant domain for the primary function must be determined from the physical limitations described in the problem.

• Calculus techniques, such as finding critical points and using the First or Second Derivative Test, are required to find the optimal value and, crucially, to justify that the result is a maximum or a minimum.

Step-by-Step Example 1: Maximizing Area

Problem: A farmer has 400 feet of fencing to enclose a rectangular pen that borders a straight river. No fencing is needed along the river. What are the dimensions of the pen that will have the largest possible area?

Step 1: Understand the Problem

• Goal: Maximize the area of a rectangular pen.

• Constraint: The total fencing available is 400 feet. The pen has three fenced sides.

• Let $x$ be the length of the two sides perpendicular to the river and $y$ be the length of the side parallel to the river.

Step 2: Write a Primary Equation

• The quantity to be maximized is Area, $A$.

• The primary equation is: $A=xy$

Step 3: Write a Secondary Equation

• The constraint is the amount of fencing. The perimeter of the three fenced sides is 400 feet.

• The secondary equation is: $2x+y=400$

Step 4: Express as a Single-Variable Function

• Solve the secondary equation for $y$: $y=400-2x$.

• Substitute this into the primary equation: $A(x)=x(400-2x)=400x-2x^2$.

Step 5: Determine the Domain

• The side length $x$ must be positive, so $x>0$.

• The side length $y$ must also be positive: $400-2x>0⟹400>2x⟹200>x$.

• The relevant domain is $0<x<200$.

Step 6: Use Calculus to Find the Optimal Value

• Find the derivative:$A^{\prime }(x)=\frac{d}{dx}(400x-2x^2)=400-4x$.

• Find critical points: Set $A^{\prime }(x)=0$.

  $$400-4x=0$$

  $$4x=400$$

  $$x=100$$

  This critical point is within our domain $(0,200)$.

• Justify the maximum: We will use the Second Derivative Test.

  • Find the second derivative: $A^{\prime \prime }(x)=-4$.

  • Since $A^{\prime \prime }(100)=-4<0$, the function $A(x)$ is concave down at $x=100$, which confirms that this critical point corresponds to a local maximum. Since it is the only critical point on the open interval, it is the absolute maximum.

• Find the dimensions:

  • $x=100$ feet.

  • Find $y$ using the secondary equation: $y=400-2(100)=400-200=200$ feet.

Answer: The dimensions of the pen with the largest area are 100 feet by 200 feet.

Step-by-Step Example 2: Minimizing Surface Area

Problem: A company needs to manufacture a cylindrical can with a volume of $1000{\text{ cm}}^3$. What should the radius and height of the can be to minimize the amount of material used (i.e., minimize the surface area)?

Step 1: Understand the Problem

• Goal: Minimize the surface area of a cylinder.

• Constraint: The volume of the cylinder must be $1000{\text{ cm}}^3$.

• Let $r$ be the radius and $h$ be the height of the can.

Step 2: Write a Primary Equation

• The quantity to be minimized is Surface Area, $S$. The surface area of a cylinder is the area of the top and bottom circles plus the area of the side.

• Primary equation: $S=2\pi r^2+2\pi rh$

Step 3: Write a Secondary Equation

• The constraint is the volume. The volume of a cylinder is $V=\pi r^{2}h$.

• Secondary equation: $1000=\pi r^{2}h$

Step 4: Express as a Single-Variable Function

• Solve the secondary equation for $h$: $h=\frac{1000}{\pi r^2}$.

• Substitute this into the primary equation:

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

  $$S(r)=2\pi r^2+\frac{2000}{r}$$

Step 5: Determine the Domain

• The radius $r$ must be a positive value.

• The domain is $r>0$.

Step 6: Use Calculus to Find the Optimal Value

• Find the derivative: First, rewrite $S(r)=2\pi r^2+2000r^{-1}$.

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

• Find critical points: Set $S^{\prime }(r)=0$.

  $$4\pi r-\frac{2000}{r^2}=0$$

  $$4\pi r=\frac{2000}{r^2}$$

  $$4\pi r^3=2000$$

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

  $$r=\sqrt[3]{\frac{500}{\pi }}$$

• Justify the minimum: We will use the First Derivative Test.

  • The critical point is $r\approx 5.419$.

  • Pick a test point in $(0,\sqrt[3]{500/\pi })$, such as $r=1$. $S^{\prime }(1)=4\pi -2000<0$.

  • Pick a test point in $(\sqrt[3]{500/\pi },\infty )$, such as $r=10$. $S^{\prime }(10)=40\pi -\frac{2000}{100}=40\pi -20>0$.

  • Since $S^{\prime }(r)$ changes from negative to positive at $r=\sqrt[3]{500/\pi }$, this corresponds to a local minimum. As it's the only critical point on the domain, it is the absolute minimum.

• Find the dimensions:

  • Radius: $r=\sqrt[3]{\frac{500}{\pi }}\text{ cm}$.

  • Height: $h=\frac{1000}{\pi r^2}=\frac{1000}{\pi (\sqrt[3]{500/\pi }{)}^2}=\frac{1000}{\pi (500/\pi {)}^{2/3}}\text{ cm}$.

Answer: The material is minimized when the radius is $r=\sqrt[3]{500/\pi }$ cm and the height is $h=\frac{1000}{\pi (500/\pi {)}^{2/3}}$ cm.

Using Your Calculator

While the setup and justification for optimization problems must be done analytically, a graphing calculator can be a powerful tool for finding numerical answers and verifying your work.

To find a critical point and optimal value:

1. Enter the Function: After deriving your single-variable primary equation, such as $A(x)=400x-2x^2‘,enteritinto‘Y1‘inyourcalculator.2.\ast \ast SettheWindow:\ast \ast Adjustyourwindow$[Xmin, Xmax]to match the domain you determined for the problem (e.g., $Xmin=0, $Xmax=200$).

2. Find the Maximum/Minimum Graphically:

   • Press [2nd]$then$[TRACE] to access the CALC menu.

   • Select 4:maximum or 3:minimum.

   • The calculator will prompt you for a "Left Bound," "Right Bound," and "Guess." Move the cursor to the left of the max/min and press [ENTER], then to the right and press [ENTER], and finally near the point for a guess and press [ENTER].

   • The calculator will display the coordinates of the maximum or minimum point. The x-coordinate is the location of the optimum, and the y-coordinate is the optimal value.

3. Find the Critical Point using the Derivative:

   • In Y2, enter the derivative of Y1 using the nDeriv function: Y2 = nDeriv(Y1, X, X).

   • Graph both Y1 and Y2.

   • Use the CALC menu and select 2:zero. Use this function on Y2 to find the x-intercept, which corresponds to the critical point where the derivative is zero.

Important Note: On a Free-Response Question, you must still show the analytical work. You must write down the derivative, set it equal to zero, and provide a justification (like a sign chart or second derivative test). You can state that you used a calculator to solve the equation (e.g., $A^{\prime }(x)=0⟹x\approx 5.419$), but the setup must be there.

AP Exam Quick Hit

Common Question Types

• Geometric Optimization: These are the most common type. You might be asked to find the dimensions of a shape (rectangle, cylinder, box) that maximize area or volume, or minimize perimeter or surface area, given a constraint.

  • Example: "Find the dimensions of the rectangle with maximum area that can be inscribed under the curve $f(x)=12-x^2$ in the first quadrant."

• Distance Optimization: Finding the point on a given curve that is closest to a fixed point. This involves minimizing the distance formula.

  • Example: "Find the coordinates of the point on the graph of $y=\sqrt{x}$ that is closest to the point $(4,0)$."

• Business/Economic Optimization: Finding the production level, price, or other variable that will maximize profit or revenue, or minimize cost, based on given functions.

  • Example: "The profit from selling $x$ items is given by $P(x)=100x-0.1x^2-500$. How many items must be sold to maximize profit?"

Common Mistakes

• Forgetting to Justify the Optimum: A student finds the correct critical point but fails to use the First or Second Derivative Test to prove it yields a maximum or a minimum. An answer without justification will not receive full credit.

• Answering the Wrong Question: The problem asks for the maximum area, but the student provides only the dimensions ($x$ and $y$). Or, it asks for the dimensions, and the student provides only the area. Read the prompt carefully to ensure you are providing the requested information.

• Ignoring Endpoints: For problems on a closed interval $[a,b]$, students often forget to check the function's value at the endpoints $a$ and $b$. The absolute maximum or minimum can occur at an endpoint.

• Incorrect Setup: Making an error when translating the word problem into the primary and secondary equations. For example, using the formula for the volume of a cone instead of a cylinder.

• Solving for the Wrong Variable: After finding the optimal value for one variable (e.g., $r$), forgetting to solve for the other variable ($h$) if the question asks for both dimensions.