 Hello, and welcome to a screencast today about optimization. We're going to be optimizing a population today, and as it turns out, it's a population of fish. So suppose you'd like to fish a certain lake and you know the population of perch after t months is given by the equation p of t equals 3t plus 6 sin of t plus 100. I want to know, using calculus, what is the maximum and minimum population in the first five months? Okay, well as we've been practicing a few times now, we know that we need to take the first derivative and set it equal to 0, or also see where it's undefined. So let's take a look at what that first derivative is. So the derivative, this is a pretty basic function we can do, so I see a trig function in here, so I'll have to go back and refresh that. But otherwise we're just adding, so this would be a pretty straightforward derivative to do. So the derivative of 3t is 3 plus the derivative of 6 sin of t is 6 cosine t. Then the derivative of 100, that piece goes away. Alright, so this function is obviously defined, and so we don't really have to worry about that at all, where it's undefined anyway. So all we need to do is just take this derivative and set it equal to 0, and then we're going to have to solve for t. So we can move our 3 over, so negative 3 is 6 cosine of t. Divide both sides by 6, so we get negative a half is the cosine of t. I know my parenthesis is kind of popping out here, so let me put those back in here so I'm consistent. Alright, good. So now we've got to dig back into our trig skills and figure out what the values of t are going to be. And then we're also going to need to think about what this first five months has to do with anything as well. Okay, so if you think back to your unit circle, so we know cosine is negative in the second quadrant and the third quadrant. Okay, so now you've got to think about your reference arc. So the cosine of what is going to give us half, negative a half, wherever you want to think about it. And that turns out to be pi over 3. So we know we're going to have to go pi over 3 in this direction from the axis and this direction from the axis. So in the second quadrant then we're going to end up with pi minus pi over 3, which gives me 2 pi over 3. And maybe you have these values memorized, that's fantastic too, or you got a unit circle handy, but just in case you don't, this is how you could do it again. Then in our third quadrant, we're going to do pi plus pi over 3, so that gives us a value of 4 pi over 3. Okay, fabulous. All right, so now what does it mean? Oh, and again, we've got to remember too, this goes around many, many times. So this is technically plus 2 pi k, and this is also plus 2 pi k. But going back to our limitation here, we're talking about the first five months. Okay, so obviously if you think about pi as 3.14, you know, 5 is going to be somewhere probably like over in here-ish, something a little bit less than 2 pi, but certainly greater than 3 pi over 2, I believe. But anyway, you can check that on the calculator. But yeah, so these values here, so this 2 pi k doesn't even really matter as it turns out. Okay, so this then is limiting t to be between 0 and 5. Okay, so we have to check these two values then, our two critical values that we got. So these are our CV, and then we also need to check our end points. Okay, now when I say check them, we need to go back to our original function way back here and plug all four of these values in and see which one's going to give us the biggest population and which one's going to give us the smallest population. Okay, so I want to know what p of 0 is. I want to know what p of 2 pi over 3 is. I want to know what p of 4 pi over 3 is. And then I also want to know what p of 5 is. Because the end points on here may give us a minimum or a maximum, it's just going to kind of depend on how the function works on this particular interval. Okay, so take a second plug these back into your original function and I'm going to throw some values on the screen here and make sure you guys check me on the calculator. So I believe p of 0 is 100. p of 2 pi over 3 is like 111.479. p of 4 pi over 3 is about 107.370. And you notice I'm not rounding these values too much because in case we need to compare these, you know, we want to know a few decimals anyway. And then p of 5 gives us about 109.246, let's say. Okay, so looking at these outputs then, looking at these populations, which one is the minimum and which one is the maximum? Well, the minimum is obviously 100, so that's going to give us our minimum population. So that means when you're fishing, you probably don't want to fish at the zero month whenever that happens to be. But this one here gives us our maximum at 2 pi over 3. So then that would be a certain time that would be a good time for you to go fishing because that would be when the maximum on a perch happens to be in the lake. Okay, so then we can answer this question in a sentence. Let me go down here then. So the max population of perch is 111. Oh, we can't have .4 of a fish, so let's just call that 111. 111 fish after. Okay, now 2 pi over 3. Let's see, when you figure that out, this is about 2 months. And then the minimum population of perch is 100 fish. And I guess we can put the word about in this one too because that was an approximation. It's 100 fish after zero months. Okay, so you can always graph this on your calculator or GeoJabur or some other kind of graphing utility and make sure that these numbers make sense. And then, but this is how you do it then using calculus. Thank you very much.