WEBVTT 00:00:02.000 --> 00:00:11.301 Topic 14. Optimization- Minimizing Cost and Inventory Control. We turn our attention here now to 00:00:11.303 --> 00:00:20.500 two optimization problems where we deal with construction costs and also with inventory control. 00:00:20.502 --> 00:00:29.900 We need to take a look at a model of each one and then an example. We'll start with cost. The cost function 00:00:29.902 --> 00:00:39.800 represents the total cost of enclosing a geometric figure. It could be a circle. It could be a rectangle. 00:00:39.802 --> 00:00:47.101 Our example is going to be a rectangle. We have to represent all the quantities in terms of one quantity, 00:00:47.103 --> 00:00:54.600 which we will use as x, our independent variable. This is an example of what a cost function could look like. 00:00:54.602 --> 00:01:01.301 We'll get into the example more, and you will see why it looks like it does, but you can see here that 00:01:01.303 --> 00:01:07.000 x is going to represent one of the quantities. If this is a rectangle, this would be the width, and then 00:01:07.002 --> 00:01:15.201 5,000 divided by x is going to be the length. So let's take a look. A lady wants to build a flower garden 00:01:15.203 --> 00:01:23.100 surrounded by a fence along part of her driveway. So here are the flowers. Kind of get in the mood here. 00:01:23.102 --> 00:01:30.700 You're thinking she wants to grow flowers. Ok, she wants the garden to be 5,000 square feet. So what we 00:01:30.702 --> 00:01:40.800 have here is a fixed area. She wants an attractive picket fence along the driveway side, and then she 00:01:40.802 --> 00:01:47.798 wants chain link around the rest of it. Now before we go any further, why would anybody want that? Well, 00:01:47.800 --> 00:01:54.700 because building chain link fence is a lot cheaper than a picket fence. So there's a cost involved here. 00:01:54.702 --> 00:02:02.300 And before we go to the next one let's remember that she wants it to be 5,000 square feet. So this is 00:02:02.302 --> 00:02:09.401 an idea of a visual. This is a picture of what we have in mind. The picket fence is 6 dollars a foot, 00:02:09.403 --> 00:02:16.901 and the chain link fence is 2 dollars a foot. The question is find the dimensions that will minimize the 00:02:16.903 --> 00:02:25.600 cost and then find the minimum cost. Alright, let's start with some type of picture of how this would 00:02:25.602 --> 00:02:37.798 be laid out. Ok so imagine here is maybe her driveway. So here's the street, and this is the driveway. 00:02:37.800 --> 00:02:45.101 She wants to build this enclosure along part of the driveway, and she wants the side next to the 00:02:45.103 --> 00:02:55.301 driveway to be the pretty side, so that's the picket part, and then she's going to use chain link for the 00:02:55.303 --> 00:03:01.100 rest of it. Now we don't know what the scale is here, so I'm just going to draw something that looks like 00:03:01.102 --> 00:03:18.100 a rectangle. We want this area to be, this area right here, to be 5,000 square feet, and what we 00:03:18.102 --> 00:03:26.801 want to do is minimize cost. We follow the pattern we've followed all along. We've identified what we 00:03:26.803 --> 00:03:36.997 want to do here. Now we identify the variable x. There is no real reason why this side has to be x or 00:03:36.999 --> 00:03:45.401 this side has to be x. Generally though, it is more efficient if you make the side that is different 00:03:45.403 --> 00:03:53.300 have the simpler variable. So I'm going to call this side x. So that's x feet, the length of this side. 00:03:53.302 --> 00:04:11.500 Of course that forces this side to also be x. So x is the length of the picket fence side, and our 00:04:11.502 --> 00:04:17.900 diagram here shows it's also the length of the side parallel to the picket fence. Now we have to figure out 00:04:17.902 --> 00:04:29.001 a way to represent these two sides in terms of x. The only quantity we have is area, and if you remember, 00:04:29.003 --> 00:04:40.700 the area of a rectangle is length times width. So it's going to be x times what we represent this as. 00:04:40.702 --> 00:04:51.100 Let's back up here, and let's put A of x because we want... well, let's back up here first of all and say 00:04:51.102 --> 00:05:08.000 the area is 5,000, and the area is length times width. We've used x in place of the width, so we have 00:05:08.002 --> 00:05:17.600 this relationship, and I'm going to solve now for length, and I'm going to get 5,000 divided by x. 00:05:17.602 --> 00:05:22.902 This is a standard situation that will come up in a problem like this where you have a fixed area. 00:05:22.904 --> 00:05:36.900 So we actually label this side right here as 5,000 divided by x. Let's check it. If this side is 00:05:36.902 --> 00:05:45.000 5,000 divided by x times x, what is the product? Well 5,000 divided by x, times x, the x cancels and 00:05:45.002 --> 00:05:52.701 we get 5,000, which is exactly what we want for the area. So we know how to label this side here. 00:05:52.703 --> 00:06:04.500 I'm going to rewrite it up here, just for space. That's going to be 5,000 divided by x. Now I can remove 00:06:04.502 --> 00:06:18.301 this part of the work. We want to create a model. The model is to minimize cost. So what's the cost? 00:06:18.303 --> 00:06:33.300 This side has a cost of 6 dollars. The other three have a cost of 2 dollars. So the cost as a function of x 00:06:33.302 --> 00:06:46.201 is 6 dollars times that x side plus, how much is this side? Chain link, so it's 2 dollars. This is dollars. 00:06:46.203 --> 00:06:59.798 That's feet. Dollars, feet. And then we have the length of this side which is 5,000 divided by x. We want 00:06:59.780 --> 00:07:08.500 to add the cost of this. It's 2 dollars a foot, and then there are two of them. So that would be 2 times 00:07:08.502 --> 00:07:17.600 2 times this quantity. Alright, let's check the model. 6 dollars times this number of feet, 2 dollars 00:07:17.602 --> 00:07:27.601 times this number of feet, and then two sides that are 2 dollars times that length. Let's clean up 00:07:27.603 --> 00:07:45.300 this cost function. So we get 8x plus 4 times 5, and then so that's 20 thousand over x. You can see right away 00:07:45.302 --> 00:07:50.101 that taking the derivative of this is going to be a little different than the ones we did before. 00:07:50.103 --> 00:07:55.993 All of the ones before this were maximizing profit or revenue or yield, and this one is going to be very 00:07:55.995 --> 00:08:03.000 different because we've got the x in the denominator. Ok, here we go. Model. Where is it defined? 00:08:03.002 --> 00:08:10.399 Well, it is clear that x cannot be 0 because then we'd have a denominator 0. So we're going to make this 00:08:10.401 --> 00:08:22.202 an open interval on this end. Now, how long can, if this side has to be anything greater than 0, how long can 00:08:22.204 --> 00:08:28.300 this side be? Well your first thought is well it can't be more than 5,000, but if you actually look at this, 00:08:28.302 --> 00:08:36.504 theoretically, if this side was like .0001 feet, then this side could be really long. So there's really 00:08:36.506 --> 00:08:43.902 no limit to what this side can actually be. So we end up with an open interval here that means positive reals. 00:08:43.904 --> 00:08:54.104 Alright. We have our model. Let's take the derivative. The derivative of 8x is 8. The derivative of this 00:08:54.106 --> 00:09:02.003 term, this is 20,000. This would be, I could write this as x to the negative 1. So that would be 00:09:02.005 --> 00:09:22.804 minus 20,000 x to the negative 2. So that is 8 minus 20,000 over x squared, and we want to set this equal to 0 00:09:22.806 --> 00:09:37.603 and solve for x. Let's see. Let's then have 8 equals 20,000 over x squared. We multiply both sides by 00:09:37.605 --> 00:09:48.003 x squared. Ok, so wait. This is 20,000 divided by x squared. We multiply both sides by x squared, 00:09:48.005 --> 00:09:58.604 divide both sides by 8, and take the square root of both sides. Now we get x is what? 00:09:58.606 --> 00:10:06.500 Plus or minus 50. I mean that's algebraically what we get. Now what you have to realize is that the 00:10:06.502 --> 00:10:12.602 negative 50 of course is not in this interval, so we throw that out, and we have one critical value 00:10:12.604 --> 00:10:20.601 in the interval, and it's 50. Now we have to check the second derivative to make sure that we have 00:10:20.603 --> 00:10:28.000 minimized cost. I mean the last thing you want to do is maximize cost. So we're going to have to go back 00:10:28.002 --> 00:10:34.201 here and take the function. Now let's be smart. Where would you like to take the derivative? 00:10:34.203 --> 00:10:39.299 Would you like to take the derivative of this version or this one? Well obviously this looks like the one 00:10:39.301 --> 00:10:55.102 you want to take the derivative of. So we're going to remove this part, and we check the second derivative. 00:10:55.104 --> 00:11:11.399 The second derivative would be 0, and then plus 40,000 x to the negative 3. Alright, in terms of something 00:11:11.401 --> 00:11:20.903 we can work with, this would be 40,000 over x cubed. We're going to substitute 50 into here. Now remember 00:11:20.905 --> 00:11:27.201 we don't want the actual value. We simply want to know if it's positive or negative. Positive is concave up. 00:11:27.203 --> 00:11:33.204 Negative is going to be concave down. So when you put 50 into this function you're going to find that 00:11:33.206 --> 00:11:45.202 C double prime of 50 is 40,000 over 50 cubed which is a positive number. So let's just put that this is 00:11:45.202 --> 00:11:55.498 greater than 0. Therefore, the function is concave up, and right there we have found a min. 00:11:55.500 --> 00:12:02.602 So we did indeed minimize cost. Now we go to the questions. Find the dimensions that will minimize the cost, 00:12:02.604 --> 00:12:11.901 and find the minimum cost. Well, I still think I'm going to need to do a little more work here. 00:12:11.903 --> 00:12:26.103 So let's see. I've got the critical value, which was 50. So that means that I want this side to be 50. 00:12:26.105 --> 00:12:41.200 So let's find the dimensions. That's going to be 50 feet on the picket side. Just write a little note here, 00:12:41.202 --> 00:12:53.999 and so if this one is 50, then the other side is 5,000 divided by 50, which is 100. So it's going to be 00:12:54.001 --> 00:13:04.203 50 feet by 100 feet. Alright, so we're good on the dimensions. That's what our flower garden should be. 00:13:04.205 --> 00:13:11.599 Picket fence on the 50 foot side. Check it. 50 times 100 is 5,000 square feet, so we're good on the area. 00:13:11.601 --> 00:13:21.003 Next we want the actual minimum cost. We need to go back and pull up our cost function again. So we had 00:13:21.005 --> 00:13:47.099 our cost function was 8x plus 20,000 over x, and with x as 50, we'd have 8 times 50 plus 20,000 over 50. 00:13:47.101 --> 00:13:55.401 Again you can use a calculator for this or these numbers are so friendly. 8 times 50 would be 400, and 00:13:55.403 --> 00:14:04.202 this number would be 5 into 2,000 which would be another 400. And so what is it going to cost the lady 00:14:04.204 --> 00:14:12.001 to build her pretty fence and her flower garden? 800 dollars, and that is the minimum cost. 00:14:12.003 --> 00:14:19.599 So you see we can apply the techniques of optimization to construction or even just home cost. 00:14:19.601 --> 00:14:28.599 Now let's look at another example where we apply this to inventory control. 00:14:28.601 --> 00:14:37.799 Let's look at the model for inventory control. When you're trying to control inventory, you have to think 00:14:37.801 --> 00:14:46.800 of two things. The cost function is going to be, yes, it's as always the cost of purchasing the product. 00:14:46.802 --> 00:14:54.097 In this case the example we're going to do is bags of rice. So sure you have to buy the rice. That's called 00:14:54.099 --> 00:15:00.499 the item cost, but you also have to store the rice, and that's expensive too. So you're going to have to 00:15:00.501 --> 00:15:10.000 consider two parts to this cost function. Here's an example of what a typical inventory control cost function 00:15:10.002 --> 00:15:17.099 would look like. You can see that it has lots of complex pieces. The overview here would be that this is 00:15:17.101 --> 00:15:25.900 the cost of the items and this is the cost of the storage. Now I'll stop right here, and explain to you 00:15:25.902 --> 00:15:35.500 why this is x over 2. Think of something like this. Try to imagine you order bags of rice. Let's say you 00:15:35.502 --> 00:15:43.099 order 200 at a time. They come into the store, and you have 200. You have to store them. At the end of the 00:15:43.101 --> 00:15:52.600 time that all 200 of those are sold, you had anywhere from 200 down to 0 in inventory. So over that time 00:15:52.602 --> 00:16:02.900 period the average would be 200 plus 0, divided by 2. So, in a model, of course that's theoretical. 00:16:02.902 --> 00:16:11.300 We realize it doesn't always go in perfect continuous motion. But we use the x over 2 as the average 00:16:11.302 --> 00:16:19.599 number of items in inventory at any time. So that takes care of this part. x is going to be the number 00:16:19.601 --> 00:16:28.900 of items in each order. This would be an example where we have say 5,000 bags for the year, and that represents 00:16:28.902 --> 00:16:40.099 the number of orders. Let's go to the example. A grocery store expects to sell 5,000 bags of rice this year. 00:16:40.501 --> 00:16:44.700 Should they just order all 5,000 bags in the beginning? Well, I don't know. We want to see what's going to 00:16:44.702 --> 00:16:50.901 minimize their cost. It costs 2 dollars to get the bag of rice, to buy it, and then there's a delivery 00:16:50.903 --> 00:17:00.000 charge of 50 dollars per order. Ok, now that's the item cost. It costs 2 dollars to store a bag for a year. 00:17:00.002 --> 00:17:08.200 So the question is, what should be the order size and how many times a year should the orders be placed 00:17:08.202 --> 00:17:23.099 to minimize cost? Alright, let's set up a model. This is example two. We want to minimize cost. 00:17:23.101 --> 00:17:32.198 We identify x. Now when you start to set this up you might think well x could be the number of orders 00:17:32.200 --> 00:17:40.599 or x could be the number of bags per order. It's much more efficient to let x be the number of bags 00:17:40.601 --> 00:17:57.899 of whatever items you're storing, so number of bags per order. This is called the order size. 00:17:57.901 --> 00:18:04.910 Now, if we're going to get 5,000 bags over the whole year, can you figure out how we would represent the 00:18:04.912 --> 00:18:12.710 number of orders? Well it would be the total number of bags divided by the number of bags in each order. 00:18:12.712 --> 00:18:17.810 Now that's a mouthful of the same words over and over again, but what it turns out is that the total 00:18:17.812 --> 00:18:27.910 for the year divided by the number of bags per order, and that gives you the number of orders. 00:18:27.912 --> 00:18:41.811 Alright, let's model cost. The cost will be the item cost plus the inventory cost. Ok, item cost. 00:18:41.813 --> 00:18:54.809 Every time you order rice you pay 2 dollars times the number of bags in the order, plus some kind of 00:18:54.811 --> 00:19:10.010 delivery charge, 50 dollars. That's the number of bags you order times 2 dollars plus 50. 00:19:10.012 --> 00:19:17.512 Now this is going to be multiplied, or this is going to like happen, every time you place an order. 00:19:17.514 --> 00:19:28.108 So this is the cost per order, not the cost per bag, but the cost per order, and so you would multiply that 00:19:28.110 --> 00:19:41.410 by the number of orders, which is 5,000 divided by x. Fine, we got the rice. What do we do with it? 00:19:41.412 --> 00:19:49.208 We also have to pay to store it, so let's look. It costs 2 dollars to store the bag for a year. 00:19:49.210 --> 00:19:57.710 It costs 2 dollars to store a bag. That's great. How many bags are we storing? Well we're storing 00:19:57.712 --> 00:20:05.911 however many bags are in inventory over the space of the whole year. We talked about this earlier. 00:20:05.913 --> 00:20:12.010 In these problems we will always put here x over 2. And of course I think you can probably see, 00:20:12.012 --> 00:20:17.512 this is where most of the errors come from. It's tempting here to want to put x, or maybe 5,000 over x, 00:20:17.514 --> 00:20:33.111 or some combination. This is 2 dollars, and then this is the average number of bags in inventory at any one time. 00:20:33.113 --> 00:20:39.812 One more time, why does that work? It works because we're ordering x bags at a time. Remember the example. 00:20:39.814 --> 00:20:46.709 I had 200 come in. We stored them. At the end of the time when they're sold we have 200 down to 0. The average 00:20:46.711 --> 00:20:55.612 of that would be 200 plus 0, divided by 2. Ok, we have a model. Do we have a restriction here on the domain? 00:20:55.614 --> 00:21:06.909 x cannot be 0. It would make the denominator 0. What is the total number of bags for the year? 00:21:06.911 --> 00:21:15.109 5,000. So we can't possibly have the number of bags per order be more than 5,000. We could order them all 00:21:15.111 --> 00:21:23.711 at once. So we're going to have this end a closed end. Open here and closed here. We have to order something, 00:21:23.713 --> 00:21:30.105 otherwise we'd have a 0 denominator here, but we could order all 5,000. So that's our domain. 00:21:30.107 --> 00:21:37.711 When we get a critical value it has to be in the domain. Let's see what we have. The cost function, now 00:21:37.713 --> 00:21:48.400 carefully when you multiply this. 2x times 5,000 over x, the x's cancel, so we have 2 times 5,000. We have the 00:21:48.402 --> 00:21:58.608 constant 10,000. On this term, we still have the x in the denominator. So we're going to have 5 times 5 00:21:58.610 --> 00:22:13.811 is 25 and then four zeroes. Ok, that takes care of the item cost, and then here we get x. Alright, there's the model. 00:22:13.813 --> 00:22:23.010 Let's minimize. The derivative would be, the derivative of 10,000 is 0. The derivative of this term, 00:22:23.012 --> 00:22:37.510 think of it as x to the negative 1. The derivative would be negative 250,000 x to the negative 2. 00:22:37.512 --> 00:22:46.708 Don't lose this. I've seen this happen a lot. The derivative here is 1. We set this equal to 0, and we 00:22:46.710 --> 00:23:01.615 solve for x. We have negative 250,000 all over x squared is negative 1. Negative on both sides, we can 00:23:01.617 --> 00:23:12.810 do that. Then we have 250,000 is x squared. On these problems it's kind of nice when you get to this point 00:23:12.812 --> 00:23:18.105 and you realize this number is a perfect square you probably think you did it right. In real life, it 00:23:18.107 --> 00:23:25.711 probably wouldn't be that way. You take the square root of both sides. You get x is plus or minus 500, 00:23:25.713 --> 00:23:35.407 but of course the negative is not in our domain, so we exclude that, and so we get our critical value of 00:23:35.409 --> 00:23:45.608 500, and it is in the domain. Did we minimize or maximize the cost? So we go back here, and we're going to 00:23:45.610 --> 00:23:57.211 use this little version of the derivative, and we'll take the second derivative. C double prime of x is 00:23:57.213 --> 00:24:10.205 positive 2 times 250 is 500 thousand x to the negative 3. This derivative looks very similar to the one 00:24:10.207 --> 00:24:23.208 we had in example one. So we have 500,000 over x cubed. C double prime at 500 would be a positive number 00:24:23.210 --> 00:24:29.010 over a positive number cubed, so rather than calculate it, we just say it's greater than 0. That means the 00:24:29.012 --> 00:24:42.010 function is concave up which means we have a min. Very good. Now we need to go back and answer the questions. 00:24:42.012 --> 00:24:49.911 So we've minimized inventory cost. What do we need to do? What should be the order size? Ok, which one 00:24:49.913 --> 00:25:04.700 of these is order size? x. So the order size, our critical value is 500. So 500 bags is the order size. 00:25:04.702 --> 00:25:10.506 How many times a year should the orders be placed? Number of orders per year, so that would be 00:25:10.508 --> 00:25:28.510 5,000 over 500. We just about have it. That would be 10 orders per year. 10 orders a year. 500 bags per order. 00:25:28.512 --> 00:25:36.205 That's 5,000 bags a year, and we have minimized the cost. 00:25:36.211 --> 00:25:43.111 All of the examples in this topic and the two topics before this give you just a small sense of the 00:25:43.113 --> 00:25:51.708 power you have once you know how to use the derivative to help you optimize. The next thing we do is 00:25:51.710 --> 00:25:58.330 we move on to other classes of functions and other applications that go with them.