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Lecture

# 5.5 optimization.pdf

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Department
Mathematics
Course
MAT237Y1
Professor
Dan Dolderman
Semester
Fall

Description
Calc 1 Wed 10/29/08 5.5 - Applied Max/Min Problems Okay, everything we have done so far has to do with learning about the derivative and graphing. Now is when calculus gets a little more interesting, when we go into applied problems. An example of an applied problem is a farmer comes to you for help in how to fence o▯ his land. The idea is, he only has so much fencing material to work with, but he wants as much area fenced o▯ as possible. Of course, a lot of times, problems like these are bogus anyway because i’m sure a farmer is just going to randomly ask a stranger how to solve a problem like that, and even if he did, why can’t he just spend a few extra bucks to get more materials and fence in a larger area of his land (well, okay, maybe Home Depot ran out of wood, but that still doesn’t answer the ▯rst question!)? But, silly me, we are in the fantasy world of calculus here, and in this world, many farmers just keep pounding you with these types of questions over and over until you start screaming to make it stop. Then, and only then, will they at least change the question a little to something more like: a certain area of land needs to be fenced o▯. What’s the least amount of materials that need to be used? Example: You are trying to grow a vegetable garden. Unfortuantely, Bambi loves vegetables, too. Since you believe violence is not the answer, you decide the most humane thing to do is to build a fence around this garden. Now, the garden has to be rectangular, and you only have 200 ft of fencing to deal with. What is the largest possible area your garden can be? Solution: The idea behind any applied max/min problem is that there’s always some sort of geometric ▯gure we’re dealing with. In this case, it’s a rectangle. What formulas do we know about rectangles? Well, if x is the length and y is the width, then Perimeter is P = 2x + 2y and Area is A = xy. Now let’s see what we can throw in from this problem. The only number that we are given is 200 ft of fencing. Fencing goes along the outside of the garden, ie, it’s the perimeter. This means that the perimeter, P, is 200. In other words, 200 = 2x + 2y Now, we have two equations and two variables (x and y). So, let’s solve for one of the variables! 200 = 2x + 2y 100 = x + y y = 100 ▯ x Oh, boy! Now that we have one variable in terms of the other, we throw it into the other equation! A = xy = x(100 ▯ x) = 100x ▯ x 2 But this is an applied max/min problem. Whenever you think of max/min, you should think derivative. That’s how we ▯nd the max and min of a function, after all! dA = 100 ▯ 2x dx 1 Set this guy equal to zero (we’re trying to ▯nd the relative max here) 100 ▯ 2x = 0 2x = 100 x = 50 But remember, x represents a length here, so we have to make sure it’s positive. Namely, 0 ▯ x ▯ 100. Why 100? Because you need two sides of x (one, say, going east, the other going west). So, is 50 between 0 and 100? It sure is! The last thing we need to do is check to see which value, 0, 50, or 100 gives us the largest area (we’re checking for absolute max! Remember, you need to plug in the end points!). So, based on that perimeter equation above, when x = 0; y = 100, when x = 50, y = 50, and ▯nally when x = 100, y = 0. Throwing these numbers into the area formula, we get A = 0; 2500; 0, respectively. The largest number is 2500 ft and so that’s the largest area we can enclose with 200 ft of fencing. So what steps need to be taken to do this type of problem in general? Remember, not all problems are going to be a farmer needing to fence in his yard. You’ll probably be working with many di▯erent geometric ▯gures, some in 2-d, others in 3-d. Here’s how you can solve an applied max/min problem in general: (1) Figure out which geometric ▯gure you’re dealing with and draw it!!! (2) Figure out which formulas you will be using. Most likely, you’ll need two. For example, if you’re dealing with a rectangle, you’ll probably be dealing with the formulas for Perimeter and Area. Or, if the problem involves a box, you might need the formulas for volume and surface area. (3) Use one of the formulas to express one variable in terms of another. You only want to di▯erentiate with one variable, not two (or even three)! (4) Find the absolute max or min. Don’t forget you have restrictions because we’re dealing with lengths here! They can’t be negative! Example: Here’s a famous box problem. You have a piece of cardboard that is 3 in by 8 in. By cutting out the corners of this piece of cardboard and folding up the sides, you can create a box. Maximize the volume of this box. Solution: Well, we’re dealing with a box and we need to maximize the volume. So, what’s the formula for volume? Length▯width▯height! Cutting out the corners gives the length to be 8 ▯ 2x, the width to be 3 ▯ 2x, and the height is x. Multiplying these guys together, we get (8 ▯ 2x)(3 ▯ 2x)(x) = (24 ▯ 16x ▯ 6x + 4x )(x) = 24x ▯ 22x + 4x 3 And now we di▯erentiate and set equal to zero to ▯nd the relative extrema. dV 2 dx = 24 ▯ 44x + 12x = 0 Divide everything by 4, and we get 2 3x ▯ 11x + 6 = 0 2 And throwing it into the quadratic formula, p 11 ▯ 121 ▯ 4 ▯ 3 ▯ 6 x = 2 ▯ 3 p 11 ▯ 121 ▯ 72 = 6 p 11 ▯ 49 = 6 11 ▯ 7 = 6 11 + 7 11 ▯ 7 = ; 6 6 18 4 = 6 ; 6 2 = 3; 3 So, what are our bounds? 0 ▯ x ▯ 1:5. Well, notice that 3 is out of our bounds, so we get rid of it. Checking the critical point and end poi
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