Calculus: The Fence Problem

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Optimization
Maxima/Minima
The Fence Problem Page [1 of 5]
All right, so we just saw a great example of how actually finding a maxima or minima could actually be valuable by looking at trying to maximize profit, which is something we should all try to do. I would like to take a further look at these notions of finding maxima and minima of functions.
And one thing that the fellow who edits all the videos, Jim, asked afterwards and it was a very good question. So I want to actually share the question with you because it was one of the few things he said that was actually intelligent. Anyway, the question he said was, "Now, wait a minute, you know what if we were to find a whole bunch of points that were, you know, maxima? How could there be more than one maxima?" And he drew this picture here. He drew this picture here. Don't you have a monitor by the way? He drew the picture here and he said, "Now look see here, the way I see it," this is exactly what Jim said to me. He said, "The way I see it, you see, there's a max right there," which there certainly is, and then he said, "Okay, there's a min here, who cares. But there's a max here. So, here there seems to be two maxes. So which max, you know, would you pick?" And the answer is that both of these are maximum points. And the definition of a maximum point is a point where it is a summit. It might not be the greatest mountain, it might not be the tallest mountain, but as long as you're on top of a mountain, you are at a maximum. So the actual official definition of a maximum point is a point where if you go in either direction, to your right or to your left, you are going to go downward. So in particular, a maximum is a point where the function is rising up and then after that point, it's falling down. So you have a maximum if the derivative is positive before you and then negative after you. That's a maximum.
So notice that in Jim's example, in fact, his function has two maxima, one here and one here. If you want to find the maximum value of the function, this is the highest the function gets, what you do is you take the biggest of all the maxes. So in this case, you would get this one and that's called a global max or an absolute maximum. And if there are a whole bunch of minimums, which you can easily imagine - I mean Jim could of easily have done a thing like this, then there would be a lot of minimums. One minimum is here and one minima is here. And if you want to find the absolute or global minimum, you would find the smallest of all the small ones.
I thought that was a great question that Jim asked. And I really want to emphasize that I think that first of all, you find these things by taking a look at finding candidates. And the candidates are where derivatives equals zero, because that's where that curve levels off and you have a potential maximum or minimum. So then you have to check your answer by looking to the right and left to see if you're rising or falling or falling and rising and see if you really are a max or min.
What I would like to do is illustrate this technique and this idea with a whole bunch of examples. So the first example that we're going to do is the following. By the way, I pick these examples myself and I tried to pick the examples that students for millions and millions of years, calculus students, have seen these problems again and again. In fact, the crew, some of whom had actually seen calculus said, "Gee, I remember seeing that problem when I was you know, in high school." And the truth is these problems are just classics. So I thought I would share some classics with you right now. In fact, you might see some problems similar to this in your own class or in your own book.
Before I actually state these problems and start the first one, which is the famous fencing in the farm problem, I thought I would just give you my little pitch for how I believe we should solve and resolve problems. So, in fact, right now I'd like to take a moment and give you my method for solving all your problems. And when I say all your problems, I mean all your problems. Maybe you're having a problem in another class, maybe you're having a problem that's a personal problem, maybe you're having a problem with your loved one. Well, I'm going to tell you the method; it's a three-point method, for solving any problem. In particular, you'll see how wonderfully appropriate this very simple three-point method really is for solving problems that involve mathematical issues.
Okay, step one - understand what is being asked or what you want. The first - completely identify the question. If there's a problem, make sure you understand what the problem is. This as, I know, basic and simple as it sounds is a fundamental missing element in many, many students' attempts to solve mathematical problems. They get right into it and they see a lot of words and it's complicated and very threatening. And they get nervous and they start taking derivatives and they're doing things and they never stop to really understand what it is they're trying to do. I know it sounds obvious. I know it sounds simple, but it really is important and many people don't do it. Don't move forward in a solution to the issue until you understand what the issue is. Okay.
Now once you understand what the issue is the next thing you want to do is figure out what you already know. Usually in some these mathematical problems, there will be some information that is given. You want to explicitly understand that information that's given. Understand it clearly and make sure you really see what that is. State it if you need to.
And in the last step is to just find a connection, find a relationship between that which you want and that which you know. And if you find that connection in almost all cases, if not all cases, the solution of the problem drops out automatically.
So I think this three-step method of solving any problem at all is one that I'm quite proud of, proud to endorse and one that I really invite you to take seriously. Because each of these steps - I know, you're saying, "Yeah, yeah, go on with the problem, we know this." But each of these steps actually are deeper than you may think. In fact, I will always solve the problems like this and so you will see me solve problems like this. Okay, great.
Let's now turn to these classic calculus problems. These classic calculus capers. And the first one is the famous farm-fencing problem. So, here's the famous farm-fencing problem. So we're on a farm, in fact, here is our farm. See that, isn't that a beautiful farm? Amy made this for us. Beautiful farm. This is a farmhouse; of course, this is not a farm. This is where the people live. But all over here, you see, here is where the farm animals live, a lot of pigs, some chickens. There are roosters. Look at all this stuff, a lot of farm animals. Here's a pig with a flower in his mouth. You get the idea - a lot of stuff because it's very gamey, very gamey. Okay, now you don't want your animals running all over the farm, because they'll escape. You've got to fence them in. Because let's face it, if they have a choice they wouldn't be there.
Okay, so you go out and you buy some fencing. Okay, so here's the fencing you get. And you only have this much fencing - I don't know if you can see it but I'll bend it here a little bit. It's not bendable but pretend for a second. And you have a hundred feet of fencing. In fact, let me just write that down. You have a hundred feet. Oh what the heck, I'll just write it over there. A hundred feet, there you go. So there's a hundred feet of fencing, this is a fixed length. That's all you bought was a hundred feet. And you've got to fence in, and you want to fence with a rectangular fence, so you want to build a rectangular fence. Okay, fine. Now to save some space, actually, why not use one of the walls of your farmhouse as one of the walls of the fence? So how would this look? Well, that means that you would only have to build three walls. Let me just move some of farm animals out here for a second. They're going to have a little party here. And they're going to be all over each other. Look at that, I'll have two of them kissing. Look at that. They're all very friendly on the farm.
Okay, now there are a variety ways of making the fence. Let me show you some, right here live. We're trying to build a rectangular fence remember. So, for example, you may have, in fact, let me just... You may want to fence in a rectangle that looks like that. Of course, there are other possibilities for the dimensions. For example, the dimensions could be like this. Let's try this, really fast. So there's another possibility for the dimensions. You could just barely fit in a chicken, right? You see this just barely fits in there. You could sneak in a pig, maybe, a sleeping pig. A sleeping pig can go in there, that's about it. So even though I'm still using the same amount of fencing, the perimeter over here is the same. The area is very, very tiny. In fact, you can imagine if I even made that longer and squeezed it in, it would be very, very narrow. So the area would be very, very small. Whereas, without adding any more additional fencing stuff, if I do something like this, same amount of fencing, you will notice that I can fit in a lot more things. I can fit in a couple of more pigs, maybe even squeeze in another rooster. Haven't changed the amount of fencing and yet I can put more things in.
Well, how do you want to build this rectangular fence? Obviously, you want to build a rectangular fence in a manner which is going to maximize the amount of area so that you can fit in as many farm animals as possible. So the question I now want to ask is, what are the dimensions of the rectangle that we should have, which optimize or which maximizes, the area inside the fenced-in region? The question is find the dimensions. So find how long this should be and how long this should be, we're basically - we should take our fence and break it up so that the rectangular pen that we create has the largest possible area. So that's the question and I have it written over there. Now let's try to solve it.
Okay, how are we going to solve this problem? Well, we have to figure out what the dimensions are. Actually, I'm going to move these props for a second. We'll return to them in a moment. Let me try to draw a picture that might actually allow us to capture the spirit of this. Okay, so the picture I think would look something like this. Here's the wall of the farmhouse, and we're going to build this pen, which is rectangular. We want a maximized area. All right, let's go through my three-part step for solving problems.
Step one - understand what we're being asked. Okay, so let's write that out. What we want is what? Well, I'm going to do this problem actually for you right now and you can sit back and watch. But in the problems that follow, I'm actually going to ask you to interact. But just so you get a sense of what we're doing here, I'll do this one for you. What we want is the following, we were asked to find the dimensions. So we want to find the dimensions, so we want to find dimensions of the rectangular pen that maximize this area. That is all being asked.
If you want to write this out symbolically, which I think is a good thing to do; you might want to actually label the sides of this rectangle. We can use a and b, we can use x and y. If you were here in front of me, I'd ask you what your name was and I would probably us the initials of your name. I'll just use x and y though right now. And I'll call this x, since this is sort of a x direction and I'll call this y. You can use any letters you want though. So what we want to do is we want to maximize the area and so what I want to do is write down area. So area of the rectangular is base times height. So that's going to be x times y. So what I want to do is I want to find x and find y so that this is as big as possible.
Well, how you find out where something is big? Well, what you want to do is find the maximum, you want to maximize it. You want to take a derivative that is equal to zero and solve it. Well, there's a big problem here; there's two things that are unknown. So this is a function that depends upon two unknowns. We only know how to take a derivative of things that have just one unknown. So how would we possibly take the derivative of this to find the maximum? This is a problem. So we have to stop there and say we're stuck. But that's what we want, we want to maximize this.
Okay step two - determine what you know. What is it that we know in this problem? Well, we were told two things. The first thing we were told is that the fence is going to be built along the side of our farmhouse and that the fencing we got was exactly one hundred feet long. What exactly does that mean? It means that all of this and all of this and all of this - that has to add up to one hundred feet. So let's write that down. So we know that if we add up this length, that length and that length, we get a hundred. So I could write this as y plus x plus y. So that's actually x plus two y equals and we were given a hundred feet of fencing. Hundred feet, I could write feet. I tend to drop my units, which is bad.
By the way, why shouldn't we write two x plus two y for the perimeter around a rectangle? Well, we don't need to put any fencing here, because remember we're running this along the side of a house. So there's no need to put up some fence here. So in fact, we're free to only put the fencing as I illustrated in the examples we looked at; only use three of the four sides for the fencing because that last side is going to be actually using the side of the house. So this is just y, that's one x, another y, we have two y's and one x. So that's why we only need one x here. Okay, so that's what we know.
And lastly - what we want is a connection that links these things. How can I link this piece of information and that piece of information? Well, one thing that I could do would be to solve this fact for one of the variables, either x or y, and then I could actually insert that thing into this formula and drop out one of the variables. In particular, I can actually use this to reduce this function to a function of just one variable and then I could happily take the derivative, set equal to zero to find the potential high point or low point. So the connection is going to be that I can solve for this and plug into here. Now you can either solve for y and you would get that y equals one hundred minus x all divided by two. Or you could solve for x and see that x would equal one hundred minus two y. That sounds so simple to me, so let's do that.
Let's solve this for x. It also shows another happy end, which is an interesting fun fact that I think you'll enjoy. So if I solve this for x, you can solve for either one. If you want to try on your own to solve for y and work this problem through, you get the same answer. So I see that x equals a hundred minus two y. That's just taking this and solving. Now what I can do is say, "Well, wait a minute here is an x and x equals that." So I can actually insert this in for x. Why would I ever want to do that? Well, then notice that a will only depend upon things with y. Nothing else is in there but y, and that's exactly what I want because then I can differentiate it. So, let's try that right now.
What do I see? Well, what I would see is a would equal - well, one hundred minus two y, that's the x times y. So a only depends upon one unknown, y. So I can differentiate it. And what would I differentiate it with respect to? I would differentiate it with respect to the variable that's over here. So in this problem and this is the first time you're ever going to see this, I'm actually going to not take d something delta x. I'm going to take d something dy, because this is in terms of y. Y is the independent variable. So if I take the derivative da/dy, what does that equal?
Well, you could use the Product Rule here, the product. Or you could be sort of lazy and just distribute this first and then take the derivative a la cart. In fact, let me do that because you know what? The good mathematician is the lazy mathematician. If I distribute, I would see one hundred y minus two y squared. Now if I take the derivative of that, the derivative of this with respect to y, so we're treating the y as the variable. So this is just a hundred minus the derivative of this with respect to y is four y. So that's the derivative. Remember what we're trying to do here - to find the candidates, what we have to do is we have to find the values for where the derivative equals zero.
What I would like to do now is set this equal to zero and solve. I'm actually going to remove this piece of paper and I'll put it off to the side and you can see it and I can see it. What we'll do now is take the derivative, so da/dy which equals one hundred minus four y and what I do is set it equal to zero.
I want you to know what I'm doing here is setting it equal to zero. And why? Why am I doing this? I hope this is okay and clear, what I want to do is I want to find out where that tangent line levels off. Where the slope of that line is zero. The line being horizontal might be a place where I have a maximum. Also might be a place where I have a minimum so I'm going to have to be careful. Anyway, I set it equal to zero to find candidates. So I'm going to solve this for y and if I solve this for y, I can bring the one hundred over to the other side and divide by minus four and I see that y equals twenty-five. So that's a candidate for being a max or min.
Now you can check to see if it's a max. So how would I do that? Well, we just have to listen to Jim, because Jim already noticed that there could be a lot of different maxes or there could be a min, so what we've got to do is what? What we have got to do is we've got to make sure if it's really a max, what happens beforehand? Beforehand the function should be climbing, should be gaining altitude. And then afterward, the function should be falling; it should be losing altitude. So if I see the derivative being positive before and negative afterwards, then I would know that at this point this has to be a summit. And since I only found one candidate, Jim's reservation is actually null and void. Because there's only one candidate which means there's only going to be - this is either going to be a max, a min or neither. There won't be many other maxes; there won't be any other. If there were other maxes, we would have found them in this pool of candidates. This is the only person running for office, we just don't know what the office is yet. Okay, so let's take a look and see.
How would I do that? Well, one thing we could do is draw a little number line, I guess. Think of this as a number line and put our candidate in there and let's just pick a point on this side. Pick a point on this side and see what the derivative is. Is it positive or negative? Remember if the derivative is positive, the thing is increasing. If the derivative is negative, the thing is decreasing.
So let's see what's happens. Let's pick a number smaller than twenty-five. Any number will do. How about if we pick one? If we pick one and plug it in here, what is the sign? I see a hundred minus four, which would be ninety-six and that's positive. That actually tells me that this entire region, all the values I would plug into here, would give me a positive value. It couldn't change because if it were to change, then it would have to go from being positive to negative and somewhere in between, it would have to be zero and we would have found that as a candidate when we set the sequence to zero. But this is the only candidate we found, which means that in this region, it's either going to be positive or it's going be negative but it can't be switched back and forth.
So I randomly just pick one point and see if it's positive or negative, and I saw it's positive. So what I do is I put plus signs all over here. This is the little sign chart for the derivative. The derivative is positive there. What is the derivative at twenty-five? Well, we already saw it's zero. So here it levels off. And what happens here? Well, all I have to do is pick a number that's bigger than twenty-five. You may want to pick twenty-six. Or if you're lazy like me, you pick something obscene like a million. That's bigger than twenty-five. Plug in a million here and it's easy to see that this is going to be negative. What's the actual value? I don't care. All I care is if it is positive or negative. If I put in a million here, I see a negative value; so all these values must be negative. Why? Well, if they weren't all negative then at some point I would have to change from negative to positive. That would mean that some place the derivative would have to equal zero and we've only found one candidate when we solved that. So in fact, this must be a complete picture.
What does it mean for the derivative to be positive? Well, for the derivative to be positive, our function -- our area function, must be on the rise. So it must going up. And here the derivative is negative; the function must be falling. So this is a candidate and what did we see about the candidate? That candidate must be a summit, it must be a max. This is a max. So, in fact y equals twenty-five is a maximum.
So now we found out what y is, are we done with the question? I don't know. I don't remember what the question is; it was such a long time ago. Let's turn back and ask what was it that we're after. We want to find the dimensions of the pen. So we have to find y and also x. So we're not quite done. We found y, we need find x. How do we find x? Well, this formula in fact tells me how to find x, just knowing y. I found out that y has to be twenty-five. If y has to be twenty-five, that means that x would have to be a hundred minus two times twenty-five, or one hundred minus fifty, so in fact, x would have to be one hundred minus fifty. So x would have to be fifty. So, x which would have to equal fifty. So x which would equal a hundred minus two y would equal a hundred minus fifty, which equals fifty. So the dimensions, -- so answer is that the dimension of x would have to be fifty feet and y would have to be twenty-five feet.
What is the maximum area? Well, it would be fifty times twenty-five, that is the biggest area we could possibly get. If you look back at our picture what we're seeing is the following. Notice that x is fifty and y is twenty-five. So it seems as though the x has to be twice as long as the y, which is sort of almost, how I drew it, which is sort of coincidental, I guess.
Does that answer make sense? Well, let's think about it. You see, I'm allowed to use the house here. If I'm allowed to use the house, then this direction is actually a cheap direction because I only have to put fence on this side and not on this side. I get that for free. So it doesn't make -- it's not a good use of my space to put a little teeny bit, to put a little teeny bit you see, against the wall. Because that's a free direction. So I should put a lot against here. But if I put too much against there, then my area starts to shrink down, of course. So there must be some happy medium. And the happy medium, which we discovered, was that in fact what we want is we want twice as much parallel with the farmhouse as we have sticking out. And that actually makes a little bit of sense because, as I said, this is a cheap direction. So we want to put a lot there and just a little bit here because whatever we put here we have to put on this side as well.
So we found the answer to our first classic calculus caper, the old farmhouse problem. And, by the way, notice this is dimensions you can fit in all of the farm animals very happily. Look at them all there, they fit in perfectly. They fit in perfectly. The dimensions again were fifty by twenty-five. Calculus shows the answer. Up next, another classic calculus caper. See you there.