###### John Postovit

University of North Dakota
M.Ed.,Stanford University

From over 16 years of teaching experience, he has philosophy that it takes humor, patience and understanding when teaching tough subjects.

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# Tangent Lines and Optimization

John Postovit
###### John Postovit

University of North Dakota
M.Ed.,Stanford University

From over 16 years of teaching experience, he has philosophy that it takes humor, patience and understanding when teaching tough subjects.

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[0:00:00]
Tangent lines and optimization. Well this is a topic from fairly early on in the year just after you learnt how to do derivatives. It's a really important topic. What this topic does is help you find the minimums and maximum of curves, and you can also find lines that are a tangent to a curve at any spot. So we're going to begin by doing a little bit of basic review. We'll go on to a few practise problems and then we'll go to some applications. Applications involving optimization problems.

A tangent line is just what it says, it's a line. Have a look. We've got a curve here. The thing that makes a tangent line special is that, it touches a curve only at one spot. So I'm going to pick myself a spot. There you go, so there is the tangent line and it touches just at one spot there.

Well you've done line equations since Algebra 1, and remember what it requires you have to have a slope and you have a point. Well here's a point. Whenever you do these problems, you're going to be told what the point to put it at. But we need the slope and that's where the calculus comes in.

Let's try it out. We're going to do two practise problems here. One of them is a basic problem, the other one is just one little step harder. Let's have a look.

So we're supposed to find the equation of a line that's tangent to that curve, and the problem tells us where it wants the tangent to be. Let's put that on there.

[0:02:00]
(-1, -4) okay there is the point of tangency and the tangent line goes something like that. So looking at that I think it's probably a slope of 2, 3, 4 or something like that. But we don't have to guess because we've got calculus to do this. Now, how are we going to get what we need though because remember for a line you need to have a point, you got it. But you also need a slope. I know I have heard this before, oh yeah derivatives. Derivatives are slope.

What we're going to do is the derivative of the formula we want the line to be tangent to, and then we'll see what happens. Let's see y=xÂ³-3 so the derivative of that is 3xÂ², the -3 is of course a constant and derivative of a constant is 0.

Now this is kind of funny ,slope equals a formula, that something you never had in algebra one. A slope that was a formula, you did a slope that WAs a single number. but that is because you were dealing with straight lines. This is a curve, the slope here is a certain number. The slope over here is steeper, the slope right there, not as steep. Well, let's continue.

Same problem we just found this formula for the slope, we know that it's going to go through this point and this is where you'll find the point slope formula. The one that makes the equation of any line if you have the slope and one point, well the slope will be right go in a minute.

Well this slope formula only requires me to use the x number. So the slope when x is -1 is 3 times -1Â², it's just 3. So my guess for the slope earlier on that was not so far off. Put in the formula here.

[0:04:00]
So y minus equals, I always like to do it this way because just leave blanks for what you're going to substitute in. When x is -1, y is -4, put that in and put in our slope, a little bit of simplifying and we'll have our equation. See what we've got. We've got y plus 4 equals 3x plus 3, so 3x, taking 4 from both sides, there you have it. That's the equation of the line that touches a curve at that spot. Let's go and try another one that's just a little harder.

Looks a lot harder right? That's actually not that hard a derivative to do. The other thing that makes it a little bit harder is, we don't have the point it goes through yet, we just have one of the co-ordinates. But it's not really that hard to deal with. x equals 1, it's about there and we know the x coordinate of that spot is 1 but one but we don't know the y coordinate yet. You know that the tangent line is going to go something like that. Find the y coordinate is easy enough, all you have to do is put the one into that formula. So let's try that out.

We get y=ln(1+2)-1Â². So that is y=ln{3-1). Now you could leave this as an exact form. I'm going to make a decimal approximation out of it, just to make the problem attach easier.

[0:06:00]
The decimal approximation for it is about 8.1. So we've got one of the components that we need, we have a point that it goes through. Now we have got to go find the slope, so we need the derivative of that. Now this requires the chain rule, but the derivative of x+2 is just 1. So we can affect that the the chain rule didn't happen. Derivative of this one is pretty quick to do then, you get y'=1/x+2, that's the derivative of the natural logarithm. And the derivative of the -xÂ² is -2x. We're going to go to the next slide where I've got this written out, we'll continue the problem. But we're almost done.

So there is our slope formula and we just found a moment ago that it passes through (1, 8.1), that's the (x,y) coordinates. We're going to find the slope, the slope I specifically need is the slope that one. So find y'(1) that's 1 over (1+2) minus 2 times 1 which is -5/3. Substitute in and then we're done.

Okay slope is 5/3 the reason to point slope equation again, and a little bit of simplifying let's see.

[0:08:00]
We have y= -5/3x + 5/3 and add 8.1 in both sides. Save a little bit of time and we're going to leave it right there. You have 5/3 as 1 and 2/3, 1.6 repeating, that simplifies up pretty easily.

Optimization is a really critical topic, really important. Optimization occurs all the time on the AP test and for that matter, it's actually one of the greatest uses of calculus. Optimization usually means that you're trying to find a minimum or maximum, a best solution for something. So this is kind of a trivial example, but it's good for starting out.

So the problem asked you, find two positive numbers whose products is 612 and whose sum is a minimum. You could do 1 times 612, that multiplies to 612 and then the sum is 613. Or you could do 2 times 306 multiply and it's some is 612, and it's sum is 308. Okay that's a smaller sum. We want to find the minimum, anyway we don't want to do this guessing and checking stuff. When you do optimization problems, there is a few basic steps.

First thing to do is to write the basic, primary equation. It usually involves what the problem is asking for. Whatever the problem is asking you to maximise or minimise, that's your primary equation. So in this case, it's going to have to be an equation that starts with sum is, sum equals. Sometimes you have to write and substitute as a secondary equation. So I'm going to take a secondary equation. Then you go to find the first derivative and then you substitute in 0 to find the minimum and maximum.

The reason we're doing optimization in this episode, is because tangent lines are what make this work.

[0:10:00]
Let's try this problem out. Okay product of 612, two numbers. Two numbers I need two variables, so I could say my first number is x. My second number is y. And adding them up gives you the sum, that's the primary equation. This is the one we're going to do the derivative on. But we can't yet because they are two variables. And for that matter, we're not going to be doing the derivative of that you're going to be expecting.

So I've got to get rid of the variable, that's where the secondary equation comes in. There is a bit of information that we haven't used yet. Notice there in no 612 in there, the numbers have to multiply to 612. So let's write that out. x times y equals 612. Now I have got two variables with two equations, that means I can substitute them together. I'm going to solve the second equation for y. So I can substitute away the y.

So y is 612 over x and then I'll substitute to that and then I'll have an equation that has just x in it. So here we go. The sum is x + 612 over x. So we've written our primary equation, we wrote and substituted our secondary equation. Now in order to maximise it, we need to find the place that's at a peak where the slope is 0. Let me do a quick graph of this.

If you graph this sum equation, that variable will be x. The vertical one is not y, it's the sum. You could call it x but I'm going to label it sum. Then it'll look something like this.

[0:12:00]
Right there, that's where the minimum is. Where the minimum is, that's where we're going to find the x number that we need. So our goal is to find the x coordinate of that spot. Then look, since that's the bottom it's a horizontal line. What's the slope of a horizontal line? 0.

So our next two steps again are to do the derivative and then substitute it in 0 for the slope, so we can find out what that co-ordinate is.

So we're going to do the derivative of both sides. Derivative of the sum is, let's see, derivative of that is just 1 and derivative of 612/x will be -612/xÂ². If you need to, remember you can change this 612 to 612 times x to the negative first, in order to make the derivative a little bit easier.

Now we know that this gives us this slope anywhere a long the curve, but we specifically want the slope right here. We write 0, substitute that in. All I have to do now is solve this we'll actually have two numbers. I'll add this to both sides gives me 612/xÂ² equals 1.

Multiply both sides by xÂ², so we get 612=xÂ². Do the square root of both sides, and x is the square root of 612. That's going to turn out the other number is the same because remember, we all want them to multiply to 612 and the squared of 612 times another square of 612 is 612.

[0:14:00]
So our two numbers are square roots of 612 and square root of 612. Let's try one more, again like this. "Oh! No they're back." Find the maximum volume of an open top box that can be made of 8x14 sheet. These things were probably your nemesis in pre-calc. Most pre-calc courses of these days will have some of these on there. But in pre-calc you had to find the solution by using graphing calculator. Now that we're using calculus, you've been liberated from that calculator.

Let me show you what this looks like. So yo have a sheet and you can cut as much as you want, or as little you want from a corner. And after you cut that out, you can fold it up. And you get a box, so we've got our box. And this box should have a certain volume. Now there is nothing to say that I had to cut that amount out. See the amount you cut out, that's a variable. Let me show you on here.

You can cut x off of that side and x of that side. You have to cut squares out so that the box folds up to have tops that much up. So x and x and x and x, but x is variable, you can cut away more. Let's do that. I have to cut the same all way through all the corners, otherwise they won't fold up evenly. Now fold this little guy up, and all of a sudden we've got a different box. It's cut from the same sheet but it's taller and it has a different volume.

[0:16:00]
Well, if you make one with such a tinny bit cut out you get a really flat shallow box that hardly has any volume. If you cut a lot out, you have got a tall box but it doesn't have very much volume. If you cut just the right amount out, you get a goldilocks boxes that's got the maximum volume and that's what we're going to figure out. Let's go to work.

So our constraint is that the sheet is 8 by 14, so 8 that way, 14 that way. We know that we're going to have to maximise the volume, and volume is length times width times height. We look at our box, length or width. The length and width are here and here. So if I take this 14 and cut away x from this side, and x from that side, what's left in here is 14-2x. Now if I take this one, and cut away from the 8 and x from that side and an x from that side, I've got 8-2x.

So we've got the length, we've got the width and we've got the height. Length, 14-2x, width 8-2x and the height, that's x right there. Here's the volume formula put the substitutions in to it. Length times width times height. I've already distributed, to save a little bit of time. So this is a formula where y is going to be the volume. It's going to be the volume of the shape and volume is going to have a maximum. If you were to graph this up, be careful when you're doing on your graphing calculator, because these are surprisingly tall and if you did it in a standard graphing one, it wouldn't look like anything.

[0:18:00]
So when you set it up, what you want to do is set it up so that the window of the graph is about the size of the box. Remember our original sheet of paper was 8 inches, and if you cut away x and x from this side, you can cut away more and more. But the most you can ever cut away from this side is four. Going along the 14 dimension, if you cut away more and more, you could cut as much as 7 but you can't cut that much of this side. So four is the most you can cut out. That's x and when you set up your graphing window make it four or wide. This tend to be to often one 100 or 200 or something like that.

Graph would look something like that. The peak is there and the calculus is going to find the peak for us. So let's review our steps again.

We wrote the primary equation right there, we substituted in what we needed for it. We don't need a secondary equation this time because there is only one variable. Our next step is to do the derivative, substitute zero and then we're almost done.

Derivative y', in other words the slope or the volume. That will be 12xÂ² - 88x + 112. Now at this point, it might factor. If you're on the part of the AP test where graphing calculators are allowed though, don't even waste any time trying to factor it. Don't waste any time even with the product formula. Just use your graph and calculator to find the P.

If it's on the part when you're not allowed to use a calculator, then it's going to be factorable and you'll need to do it that way. This one though doesn't factor, they usually don't and are said special made to factor out. I worked this one out a little bit in advance.

[0:20:00]
And on this one if you substitute 0 for the slope, what you're going to find is that your x values are 1.638, that's one of them, and the other one is 5.695.

We've got two solutions but there is only one solution for this problem. This one right here is nonsense, you can't cut that away from the side of the box. If you did you did be cutting away the entire sheet, there would be nothing left to make a volume with. So this is the x value that's going to give you the maximum value. The problem we'll usually ask you, to find out what the maximum volume is.

Don't stop here, it's the multiple choice. They probably would give you this as an answer to see if they can fool you into stopping a little too early. Re-read the question, it said define the volume. Maximum volume is what you get when you put this into that. So the volume at 1.638 I substituted in there earlier on and worked it out, it's about 82.98.

Tangent lines, well it's an important topic. In this episode, we've covered finding the slopes of curves at particular spots. We've covered finding the equation of tangent lines, at those particular spots. you'll often see questions on the AP test where they're going to ask you to find the equation of a tangent line at a curve, or even just the slope at a particular spot. They're really common on the multiple choice section. You may or may not see an optimization problem in the free response section.

Another important thing that's coming up in the next episode though, they always ask you to find the maximums or minimums on graphs.

[0:22:00]
Maximums and minimums on graphs happen when the slope equals 0. So tune in the next episode and you'll see us take this a step further.