Journal 1/27/16 I’m a good person. Shouldn’t that be enough for me to get into heaven? To learn about max and min numbers on closed intervals pp 329: 1, 2, 3, 4, 5 Objective Tonight’s Homework

Homework Help Let’s spend the first 10 minutes of class going over any problems with which you need help.

Notes on the Closed Interval Theorem Let’s imagine we take a snippet of a function: The critical number theorem says that as long as this function is continuous in this span, we can say something about where to find maxima or minima. - At an endpoint - There is none because the function is flat - Somewhere in between on the span

Notes on the Closed Interval Theorem Example: A 10-inch-long string is to be cut into two pieces. One of the pieces will be bent to form a square, and the other piece will be formed into a circle. Find where the string should be cut to maximize the combined area of both.

Notes on the Closed Interval Theorem Example: A 10-inch-long string is to be cut into two pieces. One of the pieces will be bent to form a square, and the other piece will be formed into a circle. Find where the string should be cut to maximize the combined area of both. x 10 – x We need to get the area of both shapes in terms of just x.

Notes on the Closed Interval Theorem x 10 – x We need to get the area of both shapes in terms of just x. Circle x = 2 π r r = x/2 π Area = π (x/2 π ) 2

Notes on the Closed Interval Theorem x 10 – x We need to get the area of both shapes in terms of just x. CircleSquare x = 2 π r4s = 10 – x r = x/2 π s = (10 – x)/4 Area = π (x/2 π ) 2 Area = ((10 – x)/4) 2

Notes on the Closed Interval Theorem x 10 – x We need to get the area of both shapes in terms of just x. CircleSquare x = 2 π r4s = 10 – x r = x/2 π s = (10 – x)/4 Area = π (x/2 π ) 2 Area = ((10 – x)/4) 2 Total Area = π (x/2 π ) 2 + ((10 – x)/4) 2

Notes on the Closed Interval Theorem Total Area = π (x/2 π ) 2 + ((10 – x)/4) 2 We can use the closed interval theorem to find the max area. This theorem tells us that it will either be an endpoint (x=0, or x=10), or will happen when f’(x) = 0.

Notes on the Closed Interval Theorem Total Area = π (x/2 π ) 2 + ((10 – x)/4) 2 We can use the closed interval theorem to find the max area. This theorem tells us that it will either be an endpoint (x=0, or x=10), or will happen when f’(x) = 0. x = 0Area = 100/16 = 6.25 x = 10Area = 100/4 π = 7.95

Notes on the Closed Interval Theorem Total Area = π (x/2 π ) 2 + ((10 – x)/4) 2 We can use the closed interval theorem to find the max area. This theorem tells us that it will either be an endpoint (x=0, or x=10), or will happen when f’(x) = 0. x = 0Area = 100/16 = 6.25 x = 10Area = 100/4 π = 7.95 f’(x) = ((1/2 π ) + (1/8))x – 5/4 0 = ((1/2 π ) + (1/8))x – 5/4 x = 4.399

Notes on the Closed Interval Theorem Total Area = π (x/2 π ) 2 + ((10 – x)/4) 2 We can use the closed interval theorem to find the max area. This theorem tells us that it will either be an endpoint (x=0, or x=10), or will happen when f’(x) = 0. x = 0Area = 100/16 = 6.25 x = 10Area = 100/4 π = 7.95 f’(x) = ((1/2 π ) + (1/8))x – 5/4 0 = ((1/2 π ) + (1/8))x – 5/4 x = 4.399(plug this back in to f(x)) x = 4.399Area = 3.5

Notes on the Closed Interval Theorem Total Area = π (x/2 π ) 2 + ((10 – x)/4) 2 We can use the closed interval theorem to find the max area. This theorem tells us that it will either be an endpoint (x=0, or x=10), or will happen when f’(x) = 0. x = 0Area = 100/16 = 6.25 x = 10Area = 100/4 π = 7.95  max area f’(x) = ((1/2 π ) + (1/8))x – 5/4 0 = ((1/2 π ) + (1/8))x – 5/4 x = 4.399(plug this back in to f(x)) x = 4.399Area = 3.5

Group Practice Look at the example problems on pages 327 and 328. Make sure the examples make sense. Work through them with a friend. Then look at the homework tonight and see if there are any problems you think will be hard. Now is the time to ask a friend or the teacher for help! pp 329: 1, 2, 3, 4, 5

Exit Question How do we define a "closed interval"? a) A segment on which our function is continuous b) Something like a circle or square where it connects c) An integral d) We don’t. “Closed interval” doesn’t e) All of the above f) None of the above