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Optimizing Complex Systems

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  • 0:40 Step 1: Visualize It
  • 1:34 Step 2: Define the Problem
  • 1:47 Step 3: Write an Equation
  • 2:22 Step 4: Find the Min/Max
  • 6:14 Step 5: Answer the Question
  • 6:48 Lesson Summary
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Lesson Transcript
Instructor: Sarah Wright
In this lesson, you'll learn how to follow a five-step process to solve complex optimization problems by visualizing, defining, writing an equation, finding the minimum or maximum, and answering the question.

Five Steps to Solve Optimization Problems

We've seen that we can solve optimization problems by following a five-step process. It is: visualize the problem, define the problem, write an equation for it, find the minimum or maximum for the problem (usually the derivatives or end-points) and answer the question.

Let's try it for a more complex problem. Take a sheet of paper that is 20 cm long by 10 cm wide. Cut a square from each corner, and fold the sheet to make an open box. How large of a square should you cut to get the largest box volume?

Visualizing the optimization problem is the first step
Visualizing a problem

Step 1: Visualize It

The first step is to visualize it. So let's draw out on a sheet of paper a rectangle that is 20 cm long by 10 cm wide. I'm going to cut a square from each corner and fold the sheet to make an open box. At this point, you may want to grab your own sheet of paper, cut the corners and fold it to make a box. I've color-coded the squares that I've cut out and the adjoining edges to create edges on the box. Okay, I've visualized it. I don't know how large each of those squares is going to be. I'm going to call those x cm. So, I've cut an x-cm square from each corner of the box. When I do that, I end up with a box that is 10 - 2x cm deep, 20 - 2x cm wide and x cm tall.

Step 2: Define the Problem

Our second step is to define the problem. What do we need to know? We want to maximize the volume. Once we have the volume maximized, we need to know how large of a square we've cut out from each corner to get that maximum volume.

Step 3: Write an Equation

So let's write an equation for it. Here's my box: It's x tall, 10 - 2x deep and 20 - 2x wide. The volume of that box is volume equals length times width times height, or v = (x)(10 - 2x)(20 - 2x). I have one equation, and I have one thing that I can change. I can change x to maximize the volume. So x is my independent variable and my volume is my dependent variable.

2.11 and 7.89 are the two critical points for the volume equation
Optimizing Complex Systems Step 4

Step 4: Find the Min/Max

I'm going to find the minimum or maximum of this equation. Let's write it out. I'm going to expand this, and I get 4x^3 - 60x^2 + 200x. All I did was multiply x by what was in the parentheses (both sets), because I don't want to deal with any parentheses when I'm taking a derivative. I want it all written out. So now I'm going to take that derivative, dv/dx, differentiate the right-hand side and I get 12x^2 - 120x + 200.

That's not so bad, but I'm trying to find the critical point of this equation. That is, where the first derivative equals 0. Let's set that to 0 and divide both sides by 4. So my critical point is where 0 = 3x^2 - 30x + 50. I don't see an easy way to factor this, so I'm going to use the quadratic formula to find what values of x solve this equation. I get x = (30 +/- the square root of (900-600)) / 6. I can expand that out to get 5 +/- (the square root of 300)/6, or 5 +/- (10 * the square root of 3)/6, which finally simplifies to 5 +/- (5 * the square root of 3)/3. I don't like fractions or square roots, so I'm going to use a calculator to write this out and find the values of x that solve this equation. That is, the values of x that are critical points for my volume equation. Those two values are x=7.89 and x=2.11.

Let's consider these two possibilities. Maybe one is a minimum and the other is a maximum. Let's look at 7.89; it's a big number. Big numbers, for whatever reason, make people jump with joy, but not here. If x=7.89, and I were to plug it in, my height would be 7.89 and the width would be 20 - 2(7.89), which would give me roughly 4. That's not so bad, but let's look at how deep the box would be. If I plug 7.89 into the equation, then I have 10 - 2(7.89), which is roughly -15 or -16, so this whole equation gives you a negative number. All of a sudden, the depth of my box is a negative, which can't work. I can't have a negative number for a box depth. So 7.89 can't be a solution. It just doesn't fit within in our constraints, so let's get rid of it.

In this example, x=2.11 is the maximum for the equation
Optimizing Complex Systems Step 4 Number Line

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