WEBVTT 00:00:02.600 --> 00:00:11.103 Topic 13. Optimization - Maximizing Yield and Area. Again we want to optimize or make the best of, 00:00:11.105 --> 00:00:19.704 but this time we want to optimize agriculture, agricultural yield, and land or land management. 00:00:19.706 --> 00:00:28.602 So let's take a look at how we would model yield. The yield function is very similar to the revenue 00:00:28.604 --> 00:00:37.801 function, except instead of money we get a product, like apples. We use y of x just to help us remember 00:00:37.803 --> 00:00:44.901 that it is yield again, but keep in mind that it is a similar setup to revenue. The yield is the total yield 00:00:44.903 --> 00:00:54.202 generated by producing and selling x units of a product at price p. So the model is y of x is the yield 00:00:54.204 --> 00:01:01.601 per unit, which may be tree or, it depends on what we're growing here, and then times the number of units. 00:01:01.603 --> 00:01:14.102 Similar to x times the price. Some examples of what a yield function might look like look a lot like 00:01:14.104 --> 00:01:22.302 a revenue function. You could just have, this could be maybe 40 apples per tree times the number of trees. 00:01:22.304 --> 00:01:29.101 Or you could have a more complicated function where you have something reducing and something increasing. 00:01:29.103 --> 00:01:43.300 Let's take a look at an example. An apple orchard contains 30 apple trees. So that's right now. It has 30 apple trees. 00:01:43.302 --> 00:01:54.800 Each tree yields about 400 apples over the growing season. So right now the yield would be 30 trees times 00:01:54.802 --> 00:02:09.702 400 apples. The owner plans to add more trees, but for each new tree, so that's each one new tree added, 00:02:09.704 --> 00:02:16.502 this would reduce the yield per tree by about 10 apples over the season. Ok, now why would that happen? 00:02:16.504 --> 00:02:24.204 Well, just think about it. If you put in more trees, the trees become more crowded. There's a fight for 00:02:24.206 --> 00:02:31.004 the nutrients, water, the sunlight, and so what happens is the yield per tree goes down. Now this is an 00:02:31.006 --> 00:02:38.600 estimate. The owner assumes that this is what's going to happen, so we use this as our model. Ok, what do we 00:02:38.602 --> 00:02:47.104 want to do? We want to know how many trees this owner should add to maximize the total yield. We want 00:02:47.106 --> 00:02:55.000 to know what the yield would be per tree, and then on this one, let's find the maximum yield for the 00:02:55.002 --> 00:03:14.602 whole orchard. Ok, first we create a model. We determine what we want to do and that is to maximize yield. 00:03:14.604 --> 00:03:27.904 We identify the variable x. The owner plans to add new trees. For each one new tree added, the yield per 00:03:27.906 --> 00:03:37.705 tree goes down by about 10 apples. So our variable here, what is changing in this problem, is each new tree. 00:03:37.707 --> 00:03:50.700 So we let x be the number of trees added. Notice that is not the total number of trees in the orchard. 00:03:50.702 --> 00:04:03.104 We already have 30. Ok, now we need a model. The model is yield, and yield is a function of x, and it is 00:04:03.106 --> 00:04:17.505 going to be the number of trees in the orchard times the yield per tree. So this is number of trees and 00:04:17.507 --> 00:04:28.801 then times the yield per tree. In doing these problems over the years, I have noticed that the students 00:04:28.803 --> 00:04:37.102 who are very successful at this and pick this up fast are the ones who actually write words down in here 00:04:37.104 --> 00:04:46.104 and then create the model. Ok, the number of trees. We have 30 now, and we want to add some number of new 00:04:46.106 --> 00:04:56.701 trees. So we're going to model this as 30 trees plus 1 and then that represents the number of trees we're 00:04:56.703 --> 00:05:09.200 going to add, and then the yield per tree is 400 right now, and it's going to go down by approximately 10. 00:05:09.202 --> 00:05:18.705 Let's see. These are trees, and so these are apples. So the number of trees and then apples per tree. 00:05:18.707 --> 00:05:34.905 That's our new function. Let's clean it up. We get 30 times 400 is 12,000, and then 400 minus 300 is 100x 00:05:34.907 --> 00:05:45.004 minus 10 x squared. It's a good model, but it probably is not going to model over the entire set of real 00:05:45.006 --> 00:05:57.304 numbers. Obviously, you can't have anything less than 0 trees added, so we know once again that 0 is the 00:05:57.306 --> 00:06:05.501 lowest number we can add. And then what happens here? This is the number of apples. Well, if x is 40, and how 00:06:05.503 --> 00:06:13.403 did I get that? I took this, set it equal to 0, solved for x. When x is 40, this number of apples per tree 00:06:13.405 --> 00:06:21.805 goes down to 0, so that would be as far as that could go. So we have an endpoint here at 40. 00:06:21.807 --> 00:06:28.503 And again, what did I do? I took the function that has the decreasing term in it, and I set it equal to 0, 00:06:28.505 --> 00:06:39.301 solved for x, and got 40. So this model is good from 0 to 40 trees added. Alright, we need a critical value. 00:06:39.301 --> 00:06:52.903 We take the derivative. We get 100 minus 20x. Set the derivative equal to 0. We get x is 5, critical value. 00:06:52.905 --> 00:07:02.907 Is the critical value in the interval? Yes, it is. Did we maximize? We check y double prime. We get 00:07:02.909 --> 00:07:12.601 negative 20 which is less than 0, so this function is concave down, and we have a max. All we have to do 00:07:12.601 --> 00:07:19.501 now is answer these questions. How many trees should be added? Well we have 30, but that wasn't 00:07:19.503 --> 00:07:30.700 the question. The question was how many should be added? So what we should do is add 5 trees. 00:07:30.702 --> 00:07:36.200 Second question. What would be the yield per tree? Now go back and look at the notes you left for yourself. 00:07:36.202 --> 00:07:42.303 Yield per tree. See how convenient this is to have it actually written down. So the yield per tree is 00:07:42.305 --> 00:07:53.504 400 minus 10x, but we're adding 5 trees, so that's 400 minus 50. So what, these aren't trees. 00:07:53.506 --> 00:07:58.800 These are apples. Let's make sure we have that straight. This is trees. This is apples. So that's 00:07:58.800 --> 00:08:10.004 400 apples minus 10 times 5, so that's 350 apples. These problems are interesting because you have so many 00:08:10.006 --> 00:08:19.104 little quantitites inside the problem. Now we need the maximum yield for the entire orchard. So we go back 00:08:19.106 --> 00:08:25.904 and we look at the actual function that models the yield. Now, you could use this version of the function, 00:08:25.906 --> 00:08:32.404 but look how quick this is going to be. We're going to go to this one right here. So we're going to say 00:08:32.406 --> 00:08:45.801 yield where the variable is 5 is going to be 30 plus 5, so that's 35 trees, which is the number of trees 00:08:45.803 --> 00:08:56.503 we're going to end up with in the orchard, times the apples per tree, which we've got as 350. 00:08:56.505 --> 00:09:09.104 Ok so we're going to multiply this together, and you can multiply this, and you're going to get 12,250 apples. That is 00:09:09.106 --> 00:09:25.801 the maximum yield for the entire orchard. So, what should we tell the owner? We need to tell him to plant 00:09:25.803 --> 00:09:36.705 5 more trees, and he will get the most, the highest, the greatest yield that he can from his orchard. 00:09:36.707 --> 00:09:47.704 Another application of this exact same concept has to do with area. So let's see how we can model area. 00:09:47.706 --> 00:09:55.104 Ok now. Think about area. There are some standard formulas for calculating the area, and we also know 00:09:55.106 --> 00:10:00.203 some things about perimeter. In particular, the easiest one to work with is a rectangle, and that's the 00:10:00.205 --> 00:10:06.503 example we're going to look at. Alright, the area function, capital A of x, represents the total area 00:10:06.505 --> 00:10:13.005 of some geometric figure. The actual formula is going to be determined by the figure. In this example, 00:10:13.007 --> 00:10:19.104 the one we're going to do, one of these represents the width and the other represents the length. So 00:10:19.106 --> 00:10:24.301 the area is length times width. It's a rectangle. Let me mention something to you that sometimes gets 00:10:24.303 --> 00:10:32.602 in the way in these problems. We tend to think of the fact that the length of a rectangle has to be longer 00:10:32.604 --> 00:10:40.404 than the width. Now those are just ideas we have on our own. There is nothing in the formula that says 00:10:40.406 --> 00:10:47.402 A equals lw means that the l and the w have to have some relationship, so when you get ready to label 00:10:47.404 --> 00:10:51.601 these quantities, you don't know what they're going to turn out to be. You just pick one to be the width 00:10:51.603 --> 00:10:58.604 and one to be the length, and whatever it turns out is fine. In this case I represented the width with x 00:10:58.606 --> 00:11:06.902 and the length here, but it could have been the other way. Let's look at an example of this. 00:11:06.904 --> 00:11:16.204 A farmer wants to make two identical rectangular enclosures along a straight river for his two young bulls. 00:11:16.206 --> 00:11:21.807 Alright, now let's get a picture of this. We want rectangular enclosures so area is going to be length 00:11:21.809 --> 00:11:26.600 times width, and the perimeter is going to be the sum of the sides. We want this along a straight river. 00:11:26.602 --> 00:11:35.201 We want identical enclosures for two young bulls. Ok, picture yourself outside. Here are the two bulls. 00:11:35.203 --> 00:11:41.102 We put them in separate pens because the bulls fight. We want them to be identical. We're along a river, 00:11:41.104 --> 00:11:48.998 and what do we want to do? Well we've got 600 yards of fence. Ok so this is maybe fence left over from 00:11:49.000 --> 00:11:54.104 somewhere else and we're going to take these pieces of fence. The question of course is how should we 00:11:54.106 --> 00:12:02.100 arrange this to get maximum area? Of course it does matter how you arrange it. So we have 600 yards of fence. 00:12:02.102 --> 00:12:13.404 Let's draw a picture of this. So here's our straight river, and we want two identical pens. Now we can't 00:12:13.406 --> 00:12:22.903 draw this to scale because that's what we're trying to find, so we're just going to draw two pens here 00:12:22.905 --> 00:12:33.698 and just think of two bulls, one in each pen. Now we need to be able to, we want to maximize area. So 00:12:33.700 --> 00:12:41.502 the first thing we always do is write down maximize area, and the next thing we do is identify the variable. 00:12:41.504 --> 00:12:48.302 Now this turns out to be sort of a challenging part. There are lots of things you can do, lot's of 00:12:48.304 --> 00:12:56.502 different things you can label. So these are the bulls right here. We could label each of these little 00:12:56.504 --> 00:13:04.101 sides a width, and then these three the length, or vice versa, but what you want to do that makes the problem 00:13:04.103 --> 00:13:12.699 simplest is whatever variable you want to solve for make that one if possible so there's just one of them. 00:13:12.701 --> 00:13:22.101 Let me show you what I mean. I'm going to call this side right here, the length l, and then I'm going to call 00:13:22.103 --> 00:13:31.199 these three sides the width, but I'm just going to go ahead and use x right now for the width. So let's 00:13:31.201 --> 00:13:40.502 write that down. x equals the width. Notice again that this looks longer than here, but we don't know the 00:13:40.504 --> 00:13:52.600 scale. So x is the width of one pen we'll call them, enclosures. So what do we have now? We have to figure 00:13:52.602 --> 00:14:01.000 out a way to represent l in terms of x. So what we're going to do is, we're going to say, ok, the sum of 00:14:01.002 --> 00:14:08.101 all these sides is something, 600 yards. So we have... let's come over here a little bit and work on this. 00:14:08.103 --> 00:14:23.302 This is 600 is going to be l plus 3 x's. So that means 600 minus 3x is l. So I can represent that side as 00:14:23.304 --> 00:14:42.302 600 minus 3x. This is a rectangle, so I model area as a function of x. It is the length times the width. 00:14:42.304 --> 00:14:52.400 Alright now let's stop and check what we have. We have 600 yards of fence. This side we're going to call the length, 00:14:52.402 --> 00:15:02.001 and that means if we use up 3 x's for these 3 sides that leaves us 600 minus the 3x. So that's the other side. 00:15:02.003 --> 00:15:08.208 So we have the product of these two is our function. Let's clean this up a little bit, and then let's 00:15:08.210 --> 00:15:20.900 attempt to maximize it. Let's put this in a little box here. Keep it out of the way. Do we have a restriction here? 00:15:20.902 --> 00:15:31.102 We certainly do. What happens when x is 200? Then this side goes down to 0. So again we have a restriction. 00:15:31.104 --> 00:15:39.404 This model only works from 0 to 200. How'd I get that? I took this function, set it equal to 0, solved 00:15:39.406 --> 00:15:49.702 for x, and got 200. Ok, multiply across. 600x minus 3 x squared, and that's our function. Pretty simple once 00:15:49.704 --> 00:15:59.100 you get the model. We're ready. Let's go for a critical value. The derivative is 600 minus 6x. 00:15:59.100 --> 00:16:07.600 Set it equal to 0. Solve for x. We get x is 10. Alright, 10 what? 10 bulls? 10 what? 00:16:07.602 --> 00:16:14.000 Well, let's see. Oh, how about 100. 00:16:14.002 --> 00:16:20.304 Go back up to the picture and look. So that means this side should be 100. All three of these should be 100. 00:16:20.306 --> 00:16:26.600 So what does that leave for l? Well, we're not quite ready to get to that part yet. We've got to check to 00:16:26.602 --> 00:16:36.500 be sure that we have maximized area. So we do A double prime is negative 6, less than 0. The function is 00:16:36.502 --> 00:16:42.204 concave down. This is the second derivative test for relative extrema, and that means that we have indeed 00:16:42.206 --> 00:16:49.701 maximized this area function which models this farmer's situation. Now we're going to answer questions. 00:16:49.703 --> 00:16:56.699 What should be the dimensions of each enclosure? We already know that this is 100, and 100, and 100. 00:16:56.701 --> 00:17:04.104 So what does that leave for this part down here? Well 100 and 100 and 100 is 300. That can go to here, 00:17:04.106 --> 00:17:17.604 from 600, so that leaves 300 for this l. Let's draw it down here. This is 100. This is 100, and this is 00:17:17.606 --> 00:17:28.299 100. So that's 3 times 100. There's 300 from the 600. That leaves 300 for this. So how much is one of 00:17:28.301 --> 00:17:40.902 these sides? 150. So what are the dimensions that each enclosure should be? 100 by 150 yards. 00:17:40.904 --> 00:17:47.904 At this point you also need, if you are the one building this, you would need something to indicate that 00:17:47.906 --> 00:17:56.800 the 100's were the sides coming out from the river, and that the 150's were the sides parallel to the river. 00:17:56.802 --> 00:18:05.500 So you might want to add something here that says parallel to the river, but of course if you were talking to 00:18:05.502 --> 00:18:10.200 the farmer you could tell him that, but if you were writing it down you would need to know that because 00:18:10.202 --> 00:18:16.004 of course it makes a difference. You're going to have three of these sides and two of these. And so, 00:18:16.006 --> 00:18:24.404 that is another application of optimization. Over this topic and the previous topic, we've done 00:18:24.406 --> 00:18:32.300 everything about maximizing. The next topic has to do with minimizing. It's a little different 00:18:32.302 --> 00:18:41.000 in the setup, but the actual concept of finding the critical value is exactly the same.