Transcription
Welcome back to Math TV with Professor V. This is part two of the video on optimization for Calculus One, and we're going to be focusing on examples that pertain to economics in particular. Okay, so just keep in mind we have to identify what quantity we're trying to optimize, write a function for it using one variable, and then take the derivative, get our critical values, and verify. So I'm just going to jump right in.
If the price charged for a candy bar is P(x) cents, then x thousand candy bars will be sold in a certain city where P(x) = 24 - x/22. How many candy bars must be sold to maximize revenue? All right, so what quantity are we trying to optimize? Revenue. And just for those of you who don't have too much econ background yet, revenue is equal to price * quantity, and that should make sense. Revenue is the total amount of money coming in, and so you take however many units you're selling—that's the quantity—and you multiply by the price. They gave us here the formula for computing price in terms of quantity x; x is being measured in thousands of candy bars technically. This right here is your demand equation. You don't need to know that to solve this problem, but that's what it is. Okay, so can we write a function representing revenue in terms of just one variable, most likely x? Sure. So revenue is R(x), and that equals price. Well, they gave me the formula right here; they gave me the price function: 24 - x/22, times the quantity, times the number of candy bars sold. The quantity is just plain old x.
Okay, so now let's distribute, clean up, that way we can write our revenue function in a way where we can take the derivative easily. So it's 24x - x²/22. I'll write that as (1/22)x². All right, here's our revenue function. Good. Okay, now we need to optimize it; we need to figure out the max revenue. Let's take our derivative. R'(x) is going to be 24 - (1/22) * 2x. So what's that going to give us? 24 - (1/11)x. Critical values come from where R'(x) = 0 and where R'(x) doesn't exist. It's a polynomial, in particular a linear function, so it exists everywhere. I'm just going to set it equal to zero. So that means 24 - (1/11)x = 0. That means (1/11)x = 24, so x = 24 * 11, which is 264. And x is in thousands, so technically that's 264,000 when we give our final answer. Now I do need to verify though that this indeed, this critical value yields the max revenue. So let's go ahead, use our second derivative test. And I'm just going to write: Verify x = 264 yields max revenue. So let's go ahead, one more derivative. Here's the first derivative, R'(x). Let's take a second derivative. R''(x). Derivative of 24, that's just a constant; derivative of -(1/11)x is -1/11. And I can see already the second derivative is always negative, which means my revenue function is concave down, so the critical value where the derivative is zero indeed is a max, so we are good to go. All right, so let's make sure we answer the question now. It says, "How many candy bars must be sold?" Okay, it's not 264; the answer is 264,000 candy bars. And you know, it's kind of funny when I first learned about this when I was studying econ, it felt backwards to me. I was like, "Really? Companies sit around and figure out how many units or items they need to sell to maximize revenue, and then based on that they'll go—this is your x, right—they'll go set the price?" That was mind-blowing to me. Maybe I was just an oblivious teen.
Okay, this next one, it's not so economy, but it just has to do with minimizing cost, so I felt like let's throw it in here. A rectangular field is to be enclosed on four sides with a fence. Fencing costs five dollars per foot for two opposite sides, so five dollars a foot here and here, and then seven dollars per foot on the other sides. They didn't say why some sides are pricier. Maybe [Music] um, I'm gonna add to the story. Okay, this is through the marshland; it's very difficult. This is easier to fence. I don't know. Find the dimensions of the field of area 870 feet squared that would be the cheapest to enclose. So what are we optimizing? We're trying to minimize cost. How do I know that? The word "cheapest," yeah. Okay, so we want to minimize cost of fencing. Okay, okay, so what's—how would we describe or come up with the formula that represents the cost of the fencing? We need some variables here. I'm going to call these sides x and these sides y. So let's see if we can come up with the cost function. Okay, so the cost, well, let's see, we've got x and another x, so we've got two x's, and however long they are, they cost five dollars a foot, right, to fence. So that's going to be 2 * x * 5; that's the cost to fence those sides. Then over here I have y and another y; I have two y's. So plus 2 * y, and each of those sides cost seven dollars a foot. So cost = 10x + 14y. Okay, remember whatever quantity we're trying to optimize, we need to take its derivative, so it needs to be a function of just one variable, and right now I've got two variables. We've got to eliminate one of them. Can you see some info we haven't used yet that will let me do that? Why, yes, they told me the area of the field is 870 feet squared. So what does that have to do with the problem? Not much, but it's just going to help us relate x and y so I can cut down on how many variables we have. We know area = x * y, so that has to be 870. And then it's up to you now. Do you want to eliminate x or do you want to eliminate y? I mean, here it would be equally easy, whatever you chose. Okay, let's just solve for y in terms of x. So y = 870/x. You could do the other way around though; it'll be fine. Okay, so now I can write cost as a function: 10x + 14 * 870/x. I know, I know, and then, oh sorry, drop my calculator. Um, C(x) is going to be 10x + I just want to make sure: 14 * 870 is 12,180, and then I'll write this as x⁻¹. All right, very good. Here's the cost function. Now let's go ahead, take the derivative so we can get our critical values. So C'(x) is going to be 10. Then remember this exponent comes down in the front, so then we'll have -12,180x⁻². Good. Now to get our critical values, usually wanna put all our variables with negative exponents in the denominator and then get a common denominator. This is very important when you do graphing problems, when you do optimization. Usually you don't need to be so precise, but I'm just going to do it so you guys keep up good habits. Okay, so here's my derivative now. Critical values come from two places: where the derivative equals zero and where the derivative does not exist. Now in this case, the derivative does not exist if x² = 0, which would mean x = 0, but that's not in the domain of the problem. Okay, so I mean, yeah, of course it would be the cheapest to enclose this field if you don't even get any fencing, but we gotta, we gotta enclose the field; we have to do what was asked of us. So we're only interested in the case where C'(x) = 0, and that happens when the numerator of your derivative is zero. So that would be 10x² - 12,180 = 0. So then we have 10x² = 12,180. x² = 1,218. x = √1218. Does that simplify at all? Let's see. I'm just gonna leave it like that. Okay, from here what we need to do is verify that that yields a min. And let me add a little page since I'm out of space here. Okay, so verify that x = √1218 yields the min cost. And again, second derivative test. So here's my first derivative. Okay, I'm going to use that version to take a second derivative. So C''(x). Derivative of 10 is going to be 0. Then when I bring this 2 down in the front, it'll be positive 2 * 12,180 = 24,360x⁻³, which is 24,360/x³. And then you could see whatever you plug in for x, as long as it's not negative, the second derivative will be positive. So I'll just say it's positive, which means our cost function is concave up at that critical value, so indeed it yields a min. Okay, then let's make sure we answer the question. They said, "Find the dimensions of the field that would be the cheapest to enclose." So we have x, so x = √1218, and then we need to find y, and if you'll remember from earlier, y =—where is it?—870/x, so y would be 870/√1218. Okay, and then, I mean, if you have to leave it as exact answers, then you would say y = 870/√1218 and x = √1218. Since we came up with x and y, you want to tell them which is which. So for us, x was at five dollars per foot, and then y was that many feet at seven dollars a foot. Okay, if you want a decimal, well then let's do it. So √1218 is going to be about 34.9 feet at five dollars a foot, and then y is about 24.9 feet at seven dollars. Okay, if you're in my class though and you have no calculator, then you just leave it like this. Good, good.
All right, last example, and this one is a little tricky um, if you have no econ background, so I'll give you the background you need right now. Find the optimum number of batches, to the nearest whole number, of an item that should be produced annually in order to—and they're telling us—minimize cost if 60,000 units are to be made, and it costs four dollars to store a unit for one year, and it costs six hundred dollars to set up the factory to produce each batch. Assume that the units of this item will be sold off throughout the year, so the cost equation will use the average cost. That's important. So let me explain what's going on because this can be very confusing. You have a factory, and you need to make sixty thousand units, sixty thousand items, over the course of a year. Should you make them six thousand at a time, ten thousand at a time, a hundred at a time? Because when you make them, okay, you don't want to make all sixty thousand at once, most likely. It costs four dollars to store the items, each item, for a year, and every time you make a batch and you start cranking out batches of items, it costs six hundred dollars just to set the factory up. So there has to be a sweet spot. You don't want to store stuff for too long, so you don't want to make all sixty thousand at the beginning of the year, right? But then you don't want to be making like such small batches because it costs 600 every single time you make a batch, so you don't want to make them in like groups of a hundred; that probably is not going to be a good idea. So it's like, how many batches, how many times should I set it up and tell the factory "it's go time"? How should I break up this 60,000? I don't know. So let x = the number of batches. Okay, and we're trying to make 60,000 units total. So how many units are in one batch? Well, 60,000/x, right? So say you make 10 batches, then in each batch you would have 6,000 because you would take 60,000/x; that's how many you're making each time. All right, now what did they tell us? We want to minimize cost; at least they told us that very explicitly. So let's figure out what costs go into all of this. There's a setup cost of each batch. How much is that? Did you catch it? Six hundred dollars to set up each batch. So if each batch we have x's, x batches, that's going to be 600x. Now what is the storage cost for each batch? This is where it gets a little tricky. It says it costs four dollars to store a unit for one year, but not all of the units are going to be stored for a whole year. The first batch might be, but then the second batch is going to be stored for what? Eleven months, ten months, depending on when you produce it. So it seems like, oh my God, there's a little too much going on right now. No, they told us that we could assume that units will be sold off throughout the year, so the cost equation will use average cost. And here's the important idea here. I have just like a little diagram for you guys to look at. Okay, when we make the assumption that the problem told us here that demand is constant, then the quantity of the stock, the quantity of your um, whatever units you're making, can be seen to deplete at a constant rate over time. Okay, so throughout the year you can assume that the units are being sold off at a constant rate like this, and when inventory reaches zero, then an order is placed, and then it just replenishes like this. Okay, so what happens is the storage cost, or the holding cost, of all of your units is calculated by finding the sum of the product of the inventory at any instant and the holding cost per unit. And so all you need to remember is you just slap a ½, you slap a ½ on there, and then you have four dollars for one unit, so four dollars times how many units are there per batch? 60,000/x. So that's the number of units in a batch, and then the ½ * 4, that's your average storage or holding cost per unit. Okay, so you can just basically cut it in half because you're assuming this constant rate of depletion for your demand equation. Okay, major, don't stress about the theory of it; just know you're going to add a ½ to whatever the cost is. This is the setup cost for the batch; that's the storage cost for the batch. Add them together, and there's your total cost. Okay, so total cost, I can write it as C(x), it equals 600x + ½ * 4, that's 2 * 60,000/x. Before we take a derivative, let's clean this up. So C(x) = 600x + 120,000x⁻¹. Now let's take a derivative. So C'(x) = 600 - 120,000x⁻². I'm not interested in where C' doesn't exist because that happens when x = 0, and that's not in the domain, so who cares? So I'm just going to set C' equal to zero. So this is a little shortcut; I haven't done this before, but I wanted you guys to see just kind of a quicker way: 120,000/x², right? And then multiply both sides by x², then you have 600x² = 120,000, and then divide by 600. That's gone, bam bam, two zeros are gone, and then you have 1200/6. So then we get x² = 200. I'm not going to put plus or minus because we're not going to have a negative number of batches. So x = √200, which is 10√2 batches. Now we need to verify that this yields indeed the minimum cost. So verify x = 10√2 yields the min cost. How do we do that, guys? Yes, second derivative test. So let's take one more derivative, right, using that version of C'. So C''(x). Derivative of 600 is zero, so bring that exponent down in the front; that would be positive 240,000x⁻³, which I can rewrite as 240,000/x³. Right? And I can tell since I'm plugging in something positive for x, then this is going to be positive, which means my cost function is concave up at that critical value, so indeed it yields a min. We are good to go. Um, so the number of batches is 10√2. In the beginning of the problem they said to round to the nearest whole number. Okay, so we can do that. So 10√2 batches is the exact answer, but if, you know, we're not really going to make 10√2 batches, we need a whole number. √2 is about 1.4, so 1.4 * 10, that's about 14 batches. Okay, there you go. So it looks like more than once a month you're going to be making a batch, right? Fourteen batches throughout the year, and there you go. So that's, that's how you solve that problem. I think the trickiest thing is coming up with the cost equation, and if you don't, you know, if you haven't taken much econ, the storage cost, the fact that you just slap a one-half, assuming, assuming the cost equation uses average cost, that was a big important factor. Okay, that concludes the video. If you're in my class, I've done all of the optimization problems on your practice exam, so you should be good to go. And give the video a thumbs up, subscribe if you haven't already, and I'll talk to you guys soon. Bye.