 Hi there, and welcome to this video where we're going to do another example of an optimization problem. And this one is fairly heavy on the trigonometry, it turns out. It's always worthwhile to get a little bit of a review of our trig when we can, and problems that involve trig are very, very common in the real world anyway, so we might as well get this really mastered. So here's a very basic problem. We have a triangle here. It's an esosceles triangle. Two of the lengths of the sides are three and three. We're not given the length of the third side, but we are given an angle between the two sides that have length three, that angle is theta. And the question is what angle theta maximizes the area of this triangle? Before we do any mathematics on this, let's just go over to Geogebra and convince ourselves that this really matters, this really is an optimization problem. I probably have a sense that changing the angle in this triangle does change the area, but I need to convince myself first of all that the area can reach a maximum value, but at some point once the angle starts getting larger, the area does not keep getting larger. So I've got this set up. Here's the triangle from the problem. It's a esosceles with side length three over here, side length three over here, and although it's not labeled this angle in between here at point A prime is theta. I've got theta set up as a slider here, and just watch the area of the polygon here as I change the theta. If I slide theta all the way back to zero, obviously there's no triangle at all, but if I start increasing the angle, it grows and grows and grows, and so right about here and it begins to drop again. Okay, so there is an angle value. It's difficult to say exactly what it is. That was the largest value I managed to get by messing with my slider, but that's not really precise. So theta around 1.6 seems to give a maximum area. It's pretty clear from looking at this that I don't need to let theta go any less than zero, and I don't really need to let theta grow any larger than pi, because theta is equal to pi. It's 180 degrees and it just flips around and becomes the same triangle again. So I'm looking for the value of theta maybe on the interval from zero to pi that maximizes the area. With that in mind and being personally convinced that the area does actually reach a maximum, let's go back to the slides now and set up the problem and do some math. Okay, so now back to the picture of my problem and we're going to try to find the angle theta that maximizes the area. Now let's work through this in an organized way so we don't lose track of things. So here I've drawn the picture that helps. So I have the triangle with the sides labeled and the angle labeled. Now I want to first of all identify the quantity that I'm trying to optimize here, and that is area. Okay, area, and I'm going to denote that with an A. A for area. Now what I want to do is look in the problem and think about any formula or relationship that I know of that would possibly tell me how to find this area, and probably the one that jumps first of mine for most people is that this is an area of a triangle and it's one-half times the base times the height. However, we haven't really put the base and the height in the picture, so let's do that real quickly here. The base is just the length of this side here on the bottom. We don't know what that length is actually. It's tempting to say three, but we don't know this is an equilateral triangle, just an isosceles triangle. So that big thing down there is B. And then H is what we get when we go to the vertex up here and drop a perpendicular down to the base, and that's what H is. So I have area in terms of base and height. This is kind of far off from what I want because I know that area changes as the angle changes, and it's a relationship between area and angle that I'm interested in. So at this point, my goal here is this is a perfectly fine formula for the area of a triangle. Don't get me wrong here, but it's not really useful in my problem because it doesn't have the angle in it. So my intermediate goal here, before any calculus can reasonably take place, is to take this area formula and convert it into a formula that has area on one side and theta on the other. Right now, theta is nowhere in sight except in the picture. So I need to go to the picture and relate the base and the height of this thing back to theta, and this is where the trigonometry comes in. What I'm going to do is focus on the, let's say the right-hand half of this triangle here. So let's just pull this triangle out, split it in half, and this is what we would have. There would be a right triangle. This side length is still 3. This side length is h. The bottom side length is half of the base. So that's b over 2. And because this is an esosceles triangle, this angle up here is also cut in half. So this angle is theta over 2. So when I snap that triangle in half along the dotted line and look at the piece that's left over, I have this little right triangle here. I think I can use this right triangle to relate b back to theta and also h back to theta. If I can do that, then I can relate everything on this right-hand side back to theta. So let's see what I can do here. Well, first of all, looking at what I know about trigonometry, I have a right triangle here and this angle up here is theta over 2. And the adjacent side to that angle is h and the hypotenuse is 3. So I know one thing, and that is that the cosine of the angle theta over 2, that's the angle in this picture, is equal to the adjacent side divided by the hypotenuse length. So I do get something here. I know that h is equal to 3 times cosine of theta divided by 2. Now, I can feed that back into the formula here and I'm one step closer to having my objective, which is to have a formula that phrases a area in terms of only theta. Now, the b is what's next here, but I think I can use some more trig to help me out with that. The same triangle would say that the sine of theta over 2, okay, sine is going with old Soca Toa, that's the opposite side divided by the hypotenuse. And so if you simplify this, this is telling me that b is equal to 6 sine theta over 2. Okay, now we're talking and I can put that right back into here. So now I actually have a nice formula that gives me area in terms of angle. It may need to be simplified a little more, but we can work with it at first here. Area is one half, instead of base, I just now found the base was equal to 6 sine theta over 2 times the height, and the height is 3 cosine theta over 2. And, of course, that simplifies to 9 sine theta over 2 times cosine theta over 2. Now, there are more simplifications we can do and I'm going to do that on the next slide. Here's where we left off. A is the area of the triangle is 9 times sine theta over 2 times cosine theta over 2. If we wanted to, we could proceed through the problem now without doing any further simplification, but there's one trig identity that we can pull out here that makes this really, really simple to work with. And it is the identity that says that for all angle values x that 2 times the sine of x times the cosine of x is equal to the sine of 2x. Now, why do I bring that up? Well, I see that I have a sine times cosine of the same value of an angle, so I think I can simplify here. If I just set x in this trig identity equal to theta over 2, what that gives me is 2 sine theta over 2 times cosine theta over 2 is equal to the sine of 2 times theta over 2. Does my making a substitution? And so what that lets me say here is that the sine of theta over 2 times the cosine of theta over 2 is divide both sides by 2 and I get 1 half. And then on the inside, these 2's cancel out and I have sine theta. So now I can make a nice big substitution right here and say that the area of my triangle is 9 over 2 sine theta. Really, really simple formula here. So now I have achieved my objective. Notice no calculus has taken place yet and that's really true for many optimization problems and most of it is just set up and trying to simplify as much as possible. However, I'm ready now to proceed with the calculus because the quantity I'm trying to optimize, that's area, is now phrased as a function of a single variable theta and that's where I wanted to be. So now what we're going to do is the following. We're going to try to find the absolute maximum value of this function, A, on the interval from 0 to pi because remember we said that the angle gets bigger than 180 degrees, it's just the same triangle, just flipped around one of the sides here. So what is the absolute maximum value of this function on that interval? Well, recall how we found global optimization. We had to find the critical values of this function inside that interval and then evaluate those along with the end points of the interval back into the original function. So now we're going to switch into calculus mode and take the derivative with respect to theta. So A prime of theta is equal to 9 halves cosine theta. That is the only calculus step that takes place in this entire problem. So it's totally easy, but we really have to whack through a lot of underbrush to get to it. Okay, so let's think about where this function has its critical values. Well, that would be where the derivative is either 0 or undefined. Now, the derivative is just 9 halves cosine theta. That's defined for all values of theta. So where is this equal to 0? Well, it's where cosine of theta is equal to 0. And that happens infinitely many times, but only once inside the interval from 0 to pi. And that's at theta equals pi over 2. So that's the one and only critical value of my area function inside the interval. So now I'm going to take it and go over and make a little table of values of pi over 2 and a, all right? And I'm going to evaluate in here my critical value, pi over 2, and then the two endpoints, 0 and pi. And a, remember, is over here, 9 halves sine theta. Now, the sine of pi over 2 is 1, and so the area there is 9 halves. The sine of 0 is 0, and so is the sine of pi. You might have noticed that. When I, back in georgibor, had a slider either at 0 or at 108 degrees, there was no, the quote unquote triangle was just a straight line. There was no area at all. And so we've actually done it. We have found that the area of this triangle is maximized when theta is equal to pi over 2, and the maximum value of the area is 9 halves. Now, let's go back to the georgibor picture and see if that works, given what we've just computed. We just made a bunch of calculations that showed that this triangle has a maximum area of 9 halves, and it happens when theta, the angle up there at a prime, is equal to pi over 2. But we should never totally trust our computations until we double check them with a physical model or something visual. So what I'm going to do here is just move the slider until I think that it maxes out the area, and it seems to max out at 4 and a half, which is 9 halves. And I note this by going to a calculator that pi over 2 is about 1.57, and given the decimal accuracy I have here on georgibor, that would round to 1.6. So this does actually fit. The visual tool here corroborates our calculation. So let's do a brief recap of this fairly complex problem here. We first of all, in any optimization problem, want to identify the quantity that we're being asked to maximize or minimize. In this case, we're being asked to maximize area. And so the next step in the problem is to pull out anything and everything we know about that quantity. And very readily here, we can identify that area is equal to 1 half base times height. Now, however, this may not be ready for calculation yet. The goal now is to phrase area, the target quantity I'm trying to optimize in terms of exactly one variable, namely the variable that is being manipulated. Here, it was easy to come up with a formula for area in terms of two variables, but first of all there are two variables, and neither one of those variables was the theta that was mentioned in the problem statement. So my immediate goal was a lot of algebra and a lot of trigonometry, eventually trying to get a formula that looks like what you see here in the blue circle, where area is on one side and an expression involving only that angle is on the other. Once we do that, we have a very easy step to take a derivative, and the goal of using the derivative is to find the global maximum and minimum values of my function on that closed interval on which it's defined. And once we do that, we have both the maximum area itself and the angle at which it's maximized. Thanks for watching.