5.11 Solving Optimization Problems

5.11 Solving Optimization Problems

In the last key topic, we began to take a look at optimization problems and even practiced a few. Let’s recap and do some more practice questions!

🔎 Understanding Optimization Problems

Optimization problems are a key aspect of real-world applications in calculus, and involve finding the maximum or minimum value of a function in applied contexts. These contexts can range from determining the dimensions for maximum volume to minimizing costs. The objective is to identify the optimal conditions that lead to an extreme value. 👾

📝 Optimization on the AP Calculus Exam

Optimization problems are presented in many formats on the AP Test— you may see them as part of multiple choice problems, and they are usually accompanied by lengthier context or story problems. You are, however, basically guaranteed to see them one way or another—which is why it’s so important to understand how to solve them!

---

🧩 How to Solve Optimization Problems

🧺 Identify the Objective Function

Begin by clearly defining the quantity you want to optimize. This is your objective function, often denoted by f(x) or f(y). For instance, if you're a farmer looking to maximize your crop yield, your objective function might be P(x) for the total profit.

💥 Establish Constraints

Consider any constraints or limitations on the variables involved. Constraints could be in the form of limitations on resources, dimensions, or any other relevant factors. In our farming example, this could be the amount of land available or a budget constraint for purchasing seeds and fertilizer.

🖊️ Formulate the Optimization Equation

Create an equation that represents the quantity you want to optimize. This is the function you aim to maximize or minimize. If you're maximizing profit, your equation might involve revenue minus costs: P(x) = R(x) − C(x).

🎯 Find Critical Points

Take the derivative of the objective function with respect to the variable of interest. Set the derivative equal to zero and solve for critical points. Remember, critical points are potential locations for maxima or minima.

🤔 Test Critical Points

Use the first or second derivative test to determine whether each critical point corresponds to a local maximum, minimum, or neither. This step ensures that you are pinpointing the desired extreme value.

🧠 Consider Endpoints

If the optimization problem involves a closed interval, evaluate the objective function at the endpoints as well. Include these results in your analysis.

---

📈 Optimization Practice Problems

Let’s put these steps into action and give two questions a try.

🏡 Problem 1: Maximizing Area of a Rectangular Garden

You have 100 meters of fencing to enclose a rectangular garden. What dimensions will maximize the enclosed area?

First, define your variables in terms of what was given to you.

• Let x be the width of the rectangle.
• The length, y, will be determined by the remaining fencing: $2x+2y=100$ or $y=50 -x.$

Since we’re asked to maximize the area, we can define the equations with the area formula of a rectangle. The area, A, of the rectangle is given by

We have our equations! Now we can find our critical points. Take the derivative of $A(x)$ and set it equal to zero:

Test this critical point using the second derivative test. This will confirm that $x=25$ is a maximum.

Because $A''(x)$ is a negative value, this confirms that $x = 25$ is a maximum.

Last but not least, interpret results!

The dimensions that maximize the area are $x = 25 \ m$ (width) and $y = 25 \ m$ (length). Good job! 👏

📦 Problem 2: Maximize the Size of a Can

A cylindrical can is to be made to contain $1000\pi$ cubic centimeters of liquid. Find the dimensions (radius and height) of the can that minimize the amount of material needed to manufacture the can.

Let’s again define our variables. Let r represent the radius of the cylinder and h represent the height of the cylinder.

Now we can define and simplify our equations. The quantity to be optimized in this problem is the surface area of the cylinder, which is the sum of the lateral surface area and the area of the two circular bases. The surface area A is given by:

The problem states that the can must contain $1000\pi$ cubic centimeters of liquid. The volume V of a cylinder is given by:

Since $V = 1000\pi = \pi r^2h$, we have $r^2h = 1000,$ which serves as the constraint equation.

Solve the constraint equation for h:

Then, substitute h into the equation for surface area, $A$:

Simplify this equation:

Find the critical points! Take the derivative of A with respect to r and set it equal to zero to find critical points:

Check the values of r at the endpoints of the feasible interval (in this case, r cannot be negative, so only consider positive values).

Determine the minimum values. Evaluate the area function A at the critical point and endpoints, and determine which one gives the minimum value. In this case, $r = \sqrt[3]{500}$ yields the minimum surface area.

Use the expression $h = \frac{1000}{r^2}$ to find the corresponding value of the height h. The answer is approximately h = 63.

You’re almost there! State the conclusion.

The dimensions that minimize the amount of material needed to manufacture the can are $r = \sqrt[3]{500}$ centimeters and $h = \frac{1000}{(\sqrt[3]{500})^2}$ , or approximately 63 centimeters.

---

🏆 Tips for Success

You made it to the end of this guide! Here are some tips for success:

• 💡 Clearly Define Variables: Ensure a clear understanding of the meaning of each variable in the problem.
• 📈 Graphical Insight: Consider graphing the function to visualize critical points and endpoints.
• 🧠 Units Matter: Pay attention to units in real-world problems. Ensure your final answer makes sense in the given context.

Happy optimizing!