#Optimization Problems

#Table of Contents

1. What is an Optimization Problem?
2. How to Solve an Optimization Problem
3. Worked Example
4. Practice Questions
5. Glossary
6. Summary and Key Takeaways

#What is an Optimization Problem?

Differentiation is about the rate of change of a function and provides a way of finding minimum and maximum values of a function. Anything that involves maximizing or minimizing a quantity can be modeled using differentiation. Examples include:

Minimizing the cost of raw materials used in manufacturing a product Finding the maximum height a football reaches when kicked

These are called

.

#How to Solve an Optimization Problem

In optimization problems, variables other than $x$, $y$, and $f$ are often used:

• $V$ is often used for volume, $S$ for surface area
• $r$ for radius if a circle, cylinder, or sphere is involved

Derivatives can still be found, but be clear about which letter is representing the independent ($x$) variable and which letter is representing the dependent ($y$) variable.

Problems often start by linking two connected quantities together, for example, volume and surface area. If more than one variable is involved, constraints will usually be given so that the quantity being optimized can be rewritten in terms of one variable.

#Steps to Solve an Optimization Problem

1. Rewrite the quantity to be optimized in terms of a single variable, using any constraints given in the question.
2. Differentiate and solve the derivative equal to zero to find the $x$-coordinate(s) of any critical points.
3. If there is more than one critical point, or you are required to justify the nature of the critical point, differentiate again.
4. Use the second derivative to determine the nature of each critical point and select the maximum or minimum point as necessary.
5. Interpret the answer in the context of the question.

#Worked Example

A large flower bed is being designed as a rectangle with a semicircle on each end. The total area of the bed is to be $100\pi$ meters squared.

#(a) Show that the perimeter of the bed is given by the formula $P = \pi (r + \frac{100}{r})$

Answer:

1. The width of the rectangle is $2r$ meters, and its length is $L$ meters.
2. The area consists of a semicircle, plus a rectangle, plus another semicircle:

$$\frac{1}{2}\pi r^2 + 2rL + \frac{1}{2}\pi r^2 = 100\pi$$

3. Simplify and write $L$ in terms of $r$:

$$\pi r^2 + 2rL = 100\pi \\ 2rL = 100\pi - \pi r^2 \\ L = \frac{50\pi}{r} - \frac{\pi}{2}r$$

4. The perimeter of the flower bed consists of two semicircular arcs and two straight lengths:

$$P = \pi r + \pi r + 2L \\ P = \pi r + \pi r + 2\left(\frac{50\pi}{r} - \frac{\pi}{2}r\right) \\ P = 2\pi r + \frac{100\pi}{r} - \pi r \\ P = \pi r + \frac{100\pi}{r} \\ P = \pi (r + \frac{100}{r})$$

#(b) Find the value of $r$ that minimizes the perimeter, and find the minimum value of the perimeter.

Answer:

To find the minimum value of $r$:

1. Write $P$ using the answer from part (a):

$$P = \pi (r + \frac{100}{r}) \\ P = \pi r + 100\pi r^{-1}$$

2. Differentiate $P$ with respect to $r$:

$$\frac{dP}{dr} = \pi - 100\pi r^{-2}$$

3. Set the derivative equal to zero to find critical points:

$$\pi - 100\pi r^{-2} = 0 \\ 1 - \frac{100}{r^2} = 0 \\ 1 = \frac{100}{r^2} \\ r^2 = 100 \\ r = \pm 10$$

Since $r$ is a length, we ignore the negative value:

$$r = 10$$

4. Find the minimum value of the perimeter by substituting $r$ into $P$:

$$P = \pi \left(10 + \frac{100}{10}\right) = 20\pi$$

So, $r = 10$ meters minimizes the perimeter, and the minimum value of the perimeter is $20\pi$ meters.

#(c) Justify that this is the minimum perimeter.

Answer:

To prove it is a minimum, show that the second derivative is positive (indicating the graph is concave up) at this point:

1. Find the second derivative:

$$\frac{dP}{dr} = \pi - 100\pi r^{-2} \\ \frac{d^2P}{dr^2} = 200\pi r^{-3}$$

2. Substitute $r = 10$:

$$\left.\frac{d^2P}{dr^2}\right|_{r=10} = 200\pi (10)^{-3} = \frac{200\pi}{1000} = \frac{\pi}{5} > 0$$

Note that we don't need to check endpoints here: as $r \to 0$ the perimeter function becomes unbounded, while as $r \to \infty$ the perimeter also goes to infinity.

Therefore, the second derivative is positive at $r = 10$, confirming that $20\pi$ meters is the minimum value for the perimeter.

#Practice Questions

Practice Question

Question 1: A cylindrical can is to be made to hold 1 liter of oil. Find the dimensions that will minimize the cost of the metal to make the can.

Practice Question

Question 2: Find the dimensions of a rectangle with a fixed perimeter of 20 meters that will maximize its area.

Practice Question

Question 3: A company wants to design a box with a square base and an open top. The surface area of the box is to be 108 square inches. What dimensions will maximize the volume of the box?

#Glossary

• Optimization Problem: A problem that involves finding the maximum or minimum value of a function.
• Critical Point: A point on a graph where the derivative is zero or undefined, indicating a potential maximum or minimum.
• Second Derivative Test: A method used to determine whether a critical point is a maximum, minimum, or saddle point by evaluating the second derivative at that point.
• Constraint: A condition that must be satisfied in an optimization problem.

#Summary and Key Takeaways

#Summary

Optimization problems involve finding the maximum or minimum values of a function, often subject to certain constraints. These problems can be solved using differentiation techniques, including finding critical points and using the second derivative test to determine the nature of these points.

#Key Takeaways

• Differentiation is essential for solving optimization problems.
• Always express the quantity to be optimized in terms of a single variable.
• Use the first derivative to find critical points and the second derivative to determine their nature.
• Interpret the results in the context of the original problem.

#Exam Strategy

#Real-World Applications

Optimization problems are widely applicable in various fields, including economics (e.g., maximizing profit), engineering (e.g., minimizing material use), and physics (e.g., finding optimal angles for projectile motion).

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By following these guidelines, you will be well-prepared to tackle optimization problems in your exams and in real-world scenarios. Happy studying!