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Remember that optimization problems are everywhere, but we have a five-step way to solve optimization problems. We must visualize the problem, define the problem, write an equation for the problem, find the minimum or maximum for the problem and then answer the question. With these steps, we can solve most optimization problems.
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Let's try one. What is the maximum amount of land that you can enclose in a rectangular pen that has a perimeter of 100m?
The first thing we need to do is visualize it. Let's draw this out; I have a rectangular pen. Let's say it's going to hold my turtle (he needs a lot of room). Here's my rectangular pen: it's got a height of h, a width of w and a perimeter of 100m. I don't know what h and w are since they weren't given to me. Instead, I get to pick those to solve this problem.
Next is the second step, define the problem. I need to maximize the pen area. I'm constrained by my perimeter which has to equal 100m. I don't have any other constraints, so I need to maximize this area which is really the product of my width and height of the pen.
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Let's write this as an equation - step three. Area is height times the width, but I don't know what the height or the width is. I do know that the perimeter of this rectangle is 2(height) + 2(width), and the perimeter has to be 100m. When we plug that in, we get 100 = 2h + 2w. Again, area = hw. I'm almost there, but I really want one equation. I don't know how to optimize two equations. Besides, I have three unknowns here. I'm trying to maximize the area, but I've got h and w that can change.
Let's get rid of one of those. Let's solve the perimeter equation for width. So, I've got 100 = 2h + 2w, which I can write as 50 = h + w (I've just divided everything by 2). I can solve that by subtracting h from both sides and I end up with w = 50 - h. Now I can plug my width into my area equation, so that my area just depends on the height. Now, I've got one equation, A = h(50 - h), where h is my height and A is my area.
Find the minimum or maximum is step four of our optimization process. So, we'll write out the area, A = h(50 - h), or A = 50h - hˆ2. I'm going to find the derivative of the area with respect to h. When I differentiate both sides, I get dA/dh = the derivative of the right hand side with respect to h, which is 50 - 2h. Okay, then I can find the critical point of this function by setting that derivative equal to zero (dA/dh = 0). When it's 0, h = 25.
So, if I think of my function, A is some function of h, at h = 25, my function is at a critical point. Is it a maximum or a minimum? That's a good question. Let's draw out a number line. I'll put h underneath the number line. 25 is in the middle because it's our critical point and 0 is below the h. This is actually the end of our range - we can't have a pen that has a negative height. Then, 50 is the top of our range. Our pen can't be any larger than 50m long because then we don't have enough fence to close in the pen.
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So, what is the area when h = 0? Well, 0 * (50 - 0) = 0. Okay, that area is 0. At h = 25, I have 25(50 - 25) = 625. That's much bigger than 0, so this could be a max. When h = 50, then my A = 0, because 50(50 - 50) = 0. So, h = 25 is our critical point, and at that critical point we have a maximum area, 625mˆ2.
The last step in our optimization problem is to answer the question. Again, what is the maximum amount of land that you can enclose in a rectangular pen that has a perimeter of 100m? If I have a pen whose perimeter is 100m, then I will maximize the area by having a pen which is 25m in height, 25m in width and that area (which answers the question) is 625mˆ2.
In general for optimization problems, if you follow the five-step method, you can solve most optimization problems. The hardest part is usually writing an equation, but sometimes it's in defining the problem. Once again, here are the steps:
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Math 104: Calculus16 chapters | 135 lessons | 11 flashcard sets
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