Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Optimization To optimize something means to maximize or minimize some aspect of the situation. So our steps are going to be to read the problem until we understand it, draw a picture if necessary, introduce our variables, and write an equation for the unknown quantity. Then we're going to test the critical points and endpoints of the given domain. Let's look at some examples. We want to show that among all rectangles with an 8 meter perimeter, the one with the largest area is a square. So if we have a rectangle, we know that our rectangle is made-up of length and width. So we know that two lengths +2 widths needs to equal 8. The perimeter has to equal 8. We also know that if we're maximizing the area, we have area equaling length times width. So let's solve the perimeter equation. For L, we'd get 8 -, 2 W over 2, or L would just be 4 -, W If we stick that into our A equation, we'd get 4 -, W * W or our area is 4 W minus W ^2. If we want to optimize it, we need to find the maximum or minimum, and the way we do that is we take our first derivative. So our first derivative for that area is going to be 4 - 2 W We set the first derivative equal to 0. Solving for W we get W is 2 if we know W our length was 4 -, W so our length is 2. Hence we just proved that the biggest area for an 8m perimeter is a two by two square. We could have thought about we might have had a one and a three. We could have had 1 1/2 and 3 1/2 two and two. So there were lots of possibilities that could have given us a perimeter of eight, but the biggest area would have been a two by two 2 * 2 here, 4 up here would have been 1 * 3, and here would have been 7 fourths, 617 halves times one half, 7 fourths. Yep. OK, let's look at a different 1A rectangle has its base on the X axis and its upper 2 vertices on the parabola. Y equal 12 -, X ^2. The minus X ^2 tells us that this parabola has got to be going down. So we know that Y equal 12 -, X ^2 is our parabola. So we have a rectangle that's going to look something like this where it has our two vertices on the actual parabola and the base is on the X axis. If we thought about this distance here being X, do you agree that this distance over here would also be X? Now it would be the opposite direction, so it would be negative X. So the whole distance here we could think of AS2X. So we know our base areas base times height, correct? Our base is going to be two X. We need to figure out the height of this rectangle. Well, the height of the rectangle this point is on the parabola, so this point would be XY. But we know Y is really 12 -, X ^2. Just like this point over here would be negative X and the Y would be 12 -, X ^2. So we'd have two X * 12 - X ^2. If we multiply, we get area equal 24X minus two X ^3. We're going to take the 1st derivative 24 -6 X squared. We're going to set it equal to 0 to maximize or minimize. So 6 X squared equal 24 X squared is 4X is positive or -2 So we want the dimensions of the rectangle. Well, if it's +2 and -2 we know that the base is going to be 4 and we know that the height is going to be 12 -, 2 ^2, so 12 -, 2 ^2, 12 - 4 is going to be 8. So 4 by 8 units, whatever that is. If we look at one more example or another example, what? Oh, and it asks for what is the largest area the rectangle can have? Well, so the largest area is 32 units squared. Next example, you're planning to close off a corner of the first quadrant with a line segment 20 units long running from point A zero to .0 B it's 20 long. Show that the area of the triangle enclosed by the segment is largest when A equal B. So we're trying to find the area of a triangle which we know is 1/2 base height. So area in this triangle we know that the base is going to be A and the height that's going to be B. Now can we relate A&B somehow? Well, we know it's a right triangle, so we know a ^2 + b ^2 would equal 20 ^2. So A would equal sqrt 400 -, b ^2. Substituting that in, we'd get one 1/2 square root 400 -, b ^2 * b. Now to maximize or minimize, we have to take the derivative, so a prime the derivative of this first term. So we're going to leave the half alone. The derivative of the square root is 1 / 2 sqrt 400 -, b ^2. And then by chain rule, we have to multiply by that derivative of the inside, which is going to be -2 B. And that's all multiplied by the other term B. Then we're going to add that to the derivative of the second term derivative of plain old B is 1, and we're going to multiply that by the first term so we get a prime. If we simplify this up, this 2 here and that two there are going to cancel. So we're going to get a negative b ^2 / 2 square roots of 400 -, b ^2. We want to get the second term to have the same denominator. So we're going to multiply the top and the bottom by sqrt 400 -, b ^2. So that numerator is going to turn into 400 -, b ^2. So then if we rewrite it as a single fraction, we'd get 400 - 2 B squared over 2 square roots of 400 -, b ^2. We could divide A2 out of the numerator and denominator and get 200 -, b ^2 / sqrt 400 -, b ^2. We know our critical points come when our first derivative is 0 or when our first derivative is undefined. So we get the 0 by setting the numerator equal to 0. We get the the undefined by setting the denominator. So b ^2 is 200, B would be positive and -10 square roots of two square each side we'd get b ^2 equaling 400B equal positive -20. So now we need to figure out what makes sense. If our triangle had the hypotenuse of 20, could our B be 20? Would that make a triangle? No, that doesn't make any sense, so not possible over here. We also aren't going to have negative distances, so our only distance that makes any sense is 10 square roots of two. So if we know B is 10 square roots of two, we knew that A was the square root of 20 ^2 -, b ^2, which is 10 root 2 ^2. So we'd get sqrt 400 - 200, sqrt 200, which would be 10 square roots of two. So we just proved that A equal B to get the area of the triangle enclosed by the largest area show that the area of the triangle enclosed by this segment is largest when A equal B. So we just proved that A&B are both 10 square roots of 2 units. Thank you and have a wonderful day.