# Optimization Problems

The primary idea in the business world is to maximize profit. However, it's not as simple as trying to sell as many products as possible. Other factors and costs go into a business, such as employee salaries, cost of production, cost of materials, and price of advertisement. Often, the answer to maximizing profit is not simply producing and selling as many products as possible.

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Mathematical optimization can help find the answer that maximizes profit subject to the constraints of the real world. Optimization is one of the most interesting real-world applications of Calculus. This article will further define optimization, its other applications, and a method for solving simple optimization problems.

For a focus on business and economic-type optimization problems, see our article on Applications to Business and Economics

## Optimization Problems Meaning

Mathematical optimization is the study of maximizing or minimizing a function subject to constraints, essentially finding the most effective and functional solution to a problem.

The constraints in optimization problems represent the limiting factors involved in the maximization/minimization problem. In our example of a business, the constraints would be the cost of labor, production, and advertisement. These constraints must be accounted for in our calculations as they can greatly influence the solution.

You've likely been learning and working through finding a function's extreme values (maximums and minimums). Optimization is a real-world application of finding and interpreting extreme values. Given an equation that models cost, we seek to find its minimum value, thus minimizing cost. Given an equation that models profits, we seek to find its maximum value, thus maximizing profit.

## Applications and Types of Optimization Problems

In addition to the business application we've discussed, optimization is crucial in various other fields. Optimization can be as simple as a traveler seeking to minimize transportation time. We can also apply optimization in medicine, engineering, financial markets, rational decision-making and game theory, and packaging shipments.

Optimization is also heavily discussed in computer science. Program optimization, space and time optimization, and software optimization are crucial in writing and developing efficient code and software.

Optimization problems can be quite complex, considering all the constraints involved. Converting real-world problems into mathematical models is one of the greatest challenges. As you progress through higher-level math classes, you'll deal with more complex optimization problems with more constraints to consider. In Calculus, we'll start with smaller-scale problems with fewer constraints. However, the baseline procedure is similar for all optimization problems.

## Method for Solving Optimization Problems in Calculus

Before we start working through optimization examples, we'll go through a general step-by-step method for working through these problems. Later on, we'll apply these steps as we work through real examples.

### Step 1: Fully understand the problem

Optimization problems tend to pack loads of information into a short problem. The first step to working through an optimization problem is to read the problem carefully, gathering information on the known and unknown quantities and other conditions and constraints. It may be helpful to highlight certain values within the problem.

### Step 2: Draw a diagram

To better visualize the problem, it might be helpful to draw a diagram, including labels of known values provided in the problem.

### Step 3: Introduce necessary variables

Carefully declare variable names for values that are being maximized or minimized and other unknown quantities.

### Step 4: Set up the problem by finding relationships within the problem

Use the known values and your declared variables to set up a function. You must set up your function in terms of these values and variables based on their relation to each other.

### Step 5: Find the absolute extrema

There are a couple of methods for finding absolute extrema in optimization problems.

#### The Closed Interval Method

If the domain of your function is a closed interval, the Closed Interval Method may be a good way to compute absolute extrema.

This method involves finding all critical values within the interval by setting $f\text{'}\left(x\right)=0$ and solving for $x$. Each critical point, as well as the endpoints of the interval, should be plugged in to $f\left(x\right)$. The absolute extrema are largest value and smallest value of $f\left(x\right)$ at the critical points.

#### First Derivative Test

The First Derivative Test for Absolute Extrema Values states that for a critical point $c$ of a function $f$ on an interval:

• if $f\text{'}\left(x\right)>0$ for all $x and $f\text{'}\left(x\right)<0$ for all $x>c$, then $f\left(c\right)$ is the absolute maximum value of $f\left(x\right)$

• if $f\text{'}\left(x\right)<0$ for all $x and $f\text{'}\left(x\right)>0$ for all $x>c$, then $f\left(c\right)$ is the absolute minimum value of $f\left(x\right)$

In other words, if the function goes from increasing to decreasing, it is a maximum. If the function goes from decreasing to increasing, it is a minimum.

## Constrained Optimization Problems Examples

Let's work through a common maximization problem.

### Example 1

You are tasked with enclosing a rectangular field with a fence. You are given 400 ft of fencing materials. However, there is a barn on one side of the field (thus, fencing is not required on one side of the rectangular field). What dimensions of the field will produce the largest area subject to the 400 ft of fencing materials?

We will solve this problem using the method outlined in the article.

#### Step 1: Fully understand the problem

Let's draw the important information out from the problem.

We need to fence three sides of a rectangular field such that the area of the field is maximized. However, we only have 400 ft of fencing material to use. Thus, the perimeter of the rectangle must be less than or equal to 400 ft.

#### Step 2: Draw a diagram

Clearly, you don't have to be an artist to sketch a diagram of the problem!

The diagram of the fencing problem helps us to better visualize the problem - StudySmarter Original

#### Step 3: Introduce necessary variables

Looking at the diagram above, we've already introduced some variables. We'll let the height of the rectangle be represented by $h$. We'll let the width of the rectangle be represented by $w$.

$h=height\phantom{\rule{0ex}{0ex}}w=width$

So, we can calculate area and perimeter as

$Area=h×w\phantom{\rule{0ex}{0ex}}Perimeter=h+2w$

#### Step 4: Set up the problem by finding relationships within the problem

The fencing problem wants us to maximize area $A$, subject to the constraint that the perimeter $P$ must be greater or less than 400 ft. Intuitively, we know that we should use all 400 ft of fencing to maximize the area.

So, our problem becomes:

$maximizeA=h×w\phantom{\rule{0ex}{0ex}}suchthatP=400=h+2w$

Since we seek to maximize the area, we must write the area in terms of the perimeter to achieve one single equation. In this example, we will write the area equation in terms of width, $A\left(w\right)$.

First, let's solve for the height, $h$:

$400=h+2w\phantom{\rule{0ex}{0ex}}h=400-2w$

Now, plug into the area in terms of the width equation, $A\left(w\right)$

$\begin{array}{rcl}A\left(w\right)& =& \left(400-2w\right)\left(w\right)\\ & =& 400w-2{w}^{2}\end{array}$

In this case, we solved for the variable h to write the area equation in terms of width. This is because solving for h does not yield a fractional answer, so it may be "easier" to work with for most students. It is entirely possible to solve for width and write the area equation in terms of height as well! Give it a try and see if you get the same answer!

#### Step 5: Find the absolute extrema

Now that we have a single equation containing all of the information from the problem, we want to find the absolute maximum of $A\left(w\right)$. We can define an interval for w so we can use the Closed Interval Method.

For starters, we know that w cannot be smaller than 0. If we let $h=0$, according to our perimeter equation, we have

$\begin{array}{rcl}P& =& h+2w\\ 400& =& 2w\\ w& =& 200\end{array}$

This tells us that if $h=0$, the maximum width possible is 200. So our closed interval for $w$ is $\left[0,200\right]$.

To apply the Closed Interval Method:

First, find the extrema of $A\left(w\right)$ by taking the derivative and setting it equal to 0.

$\begin{array}{rcl}A\text{'}\left(w\right)& =& 400-4w\\ 0& =& 400-4w\\ 4w& =& 400\\ w& =& 100\end{array}$

Second, plug in the critical values $w=0,w=100,andw=200intoA\left(w\right)$ and identify the largest area.

$A\left(0\right)=400\left(0\right)-2\left({0}^{2}\right)\phantom{\rule{0ex}{0ex}}=0$

$A\left(100\right)=400\left(100\right)-2\left({100}^{2}\right)\phantom{\rule{0ex}{0ex}}=20000$

$A\left(200\right)=400\left(200\right)-2\left({200}^{2}\right)\phantom{\rule{0ex}{0ex}}=0$

So, the largest value of $A$ occurs at $w=100$ where $A=20,000f{t}^{2}$.

We can confirm this using the First Derivative Test.

Graphing $A\text{'}\left(w\right)$...

We can apply the First Derivative Test to the graph of the derivative - StudySmarter Original

$A\text{'}\left(w\right)$ clearly only equals 0 at one point, $w=100$. For all $c<100$, $A\text{'}\left(w\right)$ is positive (above the x-axis). For all $c>100$, $A\text{'}\left(w\right)$ is negative (below the x-axis). So, by the First Derivative Test, $w=100$ is the absolute maximum of $A\left(w\right)$.

Let's plug in $w=100$ to our perimeter equation to find out what h should be.

$400=h+2\left(100\right)\phantom{\rule{0ex}{0ex}}h=200$

Therefore, to maximize the area enclosed by the fence subject to our material constraints, we should use a rectangle with a width of 100 ft and a height of 200 ft.

### Example 2

Now, let's try a minimization problem.

You are tasked with building a can that holds 1 liter of liquid. To maximize profit, you must build the can such that the material used to build it is minimized. What is the minimum surface area of the can required?

Again, we will solve this problem using the method outlined in the article.

#### Step 1: Fully understand the problem

Let's draw the important information out from the problem.

We need to build a can that holds 1 liter of liquid while minimizing the material used to build it. Essentially, this means we need to minimize the can's surface area.

#### Step 2: Draw a diagram

With this diagram, we can better understand what the problem is asking us to do.

The diagram of the can problem helps us to better visualize the problem - StudySmarter

#### Step 3: Introduce necessary variables

Looking at the diagram above, we've already introduced some variables. We'll let the radius of the cylindrical can be represented by $r$. We'll let the height of the cylinder be represented by $h$. So, the volume of the cylinder $V$ is $V={\mathrm{\pi r}}^{2}\mathrm{h}$ and the surface area of the cylinder $A$ is $A=2\mathrm{\pi rh}+2{\mathrm{\pi r}}^{2}$.

#### Step 4: Set up the problem by finding relationships within the problem

The can problem wants us to minimize the surface area $A$ subject to the constraint that the can must hold at least 1 liter. Intuitively, we know that to minimize surface area, we should build a can that holds 1 liter of liquid. However, since we are looking for a length measurement for $r$ and $h$, we should convert liters into cubic centimeters. Thus, we should build a can that holds 1,000 cm3 of liquid.

So, our problem becomes:

$minimizeA=2\mathrm{\pi rh}+2{\mathrm{\pi r}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{subject}\mathrm{to}V=1000={\mathrm{\pi r}}^{2}\mathrm{h}$

Since we seek to minimize the surface area, we must write the area in terms of the volume to achieve one single equation.

First, let's solve for $h$:

$1000={\mathrm{\pi r}}^{2}\mathrm{h}\phantom{\rule{0ex}{0ex}}\mathrm{h}=\frac{1000}{{\mathrm{\pi r}}^{2}}$

Now, plug into the area equation:

$A=2\mathrm{\pi r}\left(\frac{1000}{{\mathrm{\pi r}}^{2}}\right)+2{\mathrm{\pi r}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{A}=\frac{2000}{\mathrm{r}}+2{\mathrm{\pi r}}^{2}$

#### Step 5: Find the absolute extrema

Now that we have a single equation containing all the information from the problem, we want to find the absolute minimum of $A$.

We know that $r>0$. However, we do not have an upper bound for $r$.

First, we'll find the extrema of $A$ by taking the derivative and setting it equal to 0.

$A\text{'}=4\mathrm{\pi r}-\frac{2000}{{r}^{2}}\phantom{\rule{0ex}{0ex}}0=4\mathrm{\pi r}-\frac{2000}{{r}^{2}}\phantom{\rule{0ex}{0ex}}$

Graphing the derivative:

We can apply the first derivative test to the graph of the derivative - StudySmarter Original

We can see $A\text{'}=0$ at one point. We can confirm that the point $r=5.4192608391249$ is an absolute minimum for $A$ by applying the First Derivative Test. Looking at the graph, For all $c<5.4192608391249$, $A\text{'}\left(r\right)$ is negative (below the x-axis). For all $c>5.4192608391249$, $A\text{'}\left(w\right)$ is positive (above the x-axis). So, by the First Derivative Test, $r=5.4192608391249$ is the absolute maximum of $A\left(r\right)$.

Let's plug in $r=5.4192608391249$ to our volume equation to find out with $h$ should be.

$1000=\pi {\left(5.4192608391249\right)}^{2}\mathrm{h}\phantom{\rule{0ex}{0ex}}\mathrm{h}=10.8385208518578\phantom{\rule{0ex}{0ex}}$

So, to build a can that holds at least 1 liter, the minimum surface area required is

$A=2\pi \left(5.4192608391249\right)\left(10.8385208518578\right)+2\pi {\left(5.4192608391249\right)}^{2}\phantom{\rule{0ex}{0ex}}A=553.58c{m}^{2}$

## Optimization Problems - Key takeaways

• Mathematical optimization is the study of maximizing or minimizing a function subject to constraints, essentially finding the most effective and functional solution to a problem
• Optimization is a real-world application of finding and interpreting extreme values
• Solving optimization problems can seem daunting at first, but following a step-by-step procedure helps:
• Step 1: Fully understand the problem
• Step 2: Draw a diagram
• Step 3: Introduce necessary variables
• Step 4: Set up the problem by finding relationships within the problem
• Step 5: Find the absolute extrema
• To find the absolute extrema, use either the Closed Interval Method or the First Derivative Test

#### Flashcards in Optimization Problems 26

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##### Frequently Asked Questions about Optimization Problems

What is an optimization problem?

Optimization problems seek to maximize or minimize a function subject to constraints, essentially finding the most effective and functional solution to the problem.

What is an example of an optimization problem?

A real-world example of an optimization problem is the idea of maximizing profits and minimizing cost within a business.

What is the formula for solving optimization problems?

To solve an optimization problem, you must set up a function in terms of known values and variables. Then, find the extrema of the function by taking the derivative and evaluating.

What are the types of the optimization problems?

Optimization problems can be seen in a variety of fields including business, medicine, engineering, financial markets, rational decision making and game theory, packaging shipments, and computer science.

How do you determine if you are dealing with an optimization problem?

Optimization problems involve maximizing or minimizing certain quantities. To determine if a problem is an optimization problem, carefully read the problem and look for language that suggests maximizing or minimizing.

## Test your knowledge with multiple choice flashcards

Just as you would do when solving for an extreme value, to solve an optimization problem, set the _____ derivative of your equation equal to _________.

It turns out, there are some business and economic problems that you can model and solve as optimization problems in calculus. These types of problems typically involve either:

Solving business/economic optimization problems almost __ requires you to find the marginal cost or marginal revenue, and occasionally both.

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