So now we're going to look at how we use this maximum and minimum concept in say sort of a real life problem. So suppose we are given a piece of, I don't know, cardboard or whatever, and it is 10 cm by 15 cm, so it's rectangular and it's 10 by 15. And what we're going to do is we're going to cut out squares out of each corner. And this is going to then, we're going to fold the corners, fold the sides up, and this is going to make an open box, okay? And what we want to do is get the maximum volume out of the box. So what I want to know is how big to cut these little squares out, so that's what I want to know. So I'll let that be X. So that's X by X, right? That we're going to cut out. And what I want to do is maximize the volume. So the volume when I fold this up, let's see, it would be the height. So the volume is going to be a function of how big I cut out the, do the cutout and it's going to be the height, which will be X. Let's see, this length that's left in here will be 10-2x, so times that, times this length in here, which will be 15 -2s. And I'm running out of room to put this. So I may just do this, 15-2x. So there's my equation for the volume as a function of how big the cutout is, okay? So we could, let's see, we can multiply all this out. Let's see, I get V of X Is X times, what's that? 150 minus 20x minus 30x plus 4x squared, all that times x. So that's 150x, that's 50x squared plus 4x cubed. Okay, if I graphed this function, I would wind up with something that looks, let's see if I can put it where you can see it. The graph of this function is going to look something like this where it's going to come through here, it's going to come down and it's going to come back. So that's the graph of my function where this comes through looks like at about 5. But you see here and here I would have negative volumes, which doesn't make any sense. So I'm really looking at this area in here, right? And so I can look at the graph and see that's at about around 2, and this would be, the relative minimum would be about 6.5. But since I'd have a negative volume, that's, I don't want that, I want to maximize the volume. So my answer should be somewhere around 2, okay? So, that's the equation for the volume and that's a basic sketch, I always do that to see about what I should be looking at. So I'll be sure that the answer that I get makes some sense, okay? So, the next thing we're going to do to find where that actually is, what are you going to do? You're going to compute the derivative, right? So, and we're going to set the derivative equal to 0. So let's see. Let's erase some of this stuff and get some room. Okay, so let's compute the derivative of the volume. So V prime of x Is going to be 150-100x plus 12x squared. So, I want to set that equal to 0 because that would give me where the max is, right? So I set that equal to 0 and I solve for X. So let's see. I could make this, I'll make the numbers a little easier to deal with. Let's see, 12 divided by 26x squared. So, for that to be 0, this has to be 0. And I can use the quadratic formula. So, let's see, the quadratic formula. Just to remind you, X is the negative of b plus/minus the square root of b squared minus 4ac over 2a where the a is the thing in front of the x squared, the b Is the thing in front of the x and the c is the constant at the end. So, for my formula, the negative of negative 50 would be 50 plus and minus the square root of negative 50 square, which would be 2,500, minus four times that, let's see, that's 300 times that, 1,800, over twice that. So 2 times 6 is 12. So this gives me 50 plus and minus the square root of 700 over 12, which is 1.96 or 6.37. We've looked at our graph earlier and decided that was going to give me a minimum. So I didn't want that one. So that's the one I want, the 1.96. So the size of my squares that I'm going to cut out is 1.96 centimeters by 1.96 centimeters. And then I can plug that X in here to my original volume and see what volume that would give me, okay? So, that's how you can use the concept of a maximum and minimum to find the value that you need.