2. Optimization

3. Optimization (mathematics) In mathematics, the simplest case of optimization, or mathematical programming, refers to the study of problems in which one seeks to minimize or maximize a real function by systematically choosing the values of real or integer variables from within an allowed set.

4. In mathematics, maxima and minima, known collectively as extrema, are the largest value (maximum) or smallest value (minimum), that a function takes in a point either within a given neighbourhood (local extremum) or on the function domain in its entirety (global extremum).

6. We frequently encounter situations in which we are asked to do the best we can. Such a request is vague unless we are given some conditions.

7. Asking us to minimize the cost of making tables and chairs is not clear. Asking us to make the maximum number of tables and chairs possible so that the costs of production are minimized and given that the amount of material available is restricted allows us to construct a function of the situation. We can then determine the minimum or maximum of the function Such a procedure is called optimization

8. To optimize a situation is to realize the best possible outcome, subject to a set of restrictions. Because of these restrictions, the domain of the function is usually restricted. The max or min value can be determine through calculus, or might also occur at the ends of the domain.

9. and Problems

10. (p,q) Remember the graph of y = x 2 (a basic quadratic) examine the nature of the graphs (p,q)

11. If we could model a “real life situation” using a function, we could determine a maximum or minimum value for that situation ….

12. Farmer Brown has hit hard times … He and his dog Bingo can only afford 80m of fencing. He needs to maximize the area of land his cattle can feed upon. What should the dimensions of the lot be?

13. 5m 70m A = 5m X 70m = 350m 2 P = 5m + 70m + 5m = 80m

14. 8m 64 m A = 8m X 64m = 512m 2 P = 8m + 64m + 8m = 80m

15. 12m 56 m A = 12m X 56m = 672m 2 P = 12m + 56m + 12m = 80m

16. Let’s try to model this situation: 1.Maximize the Area : A 2.Let x = width, let y = length A = (w)(l) A = (x)(y) Xm Ym

17. Let’s try to model this situation: 3. The Link (second equation): P = 80 = 2w + l = 2x + y = 80 Using the 2 equations and substitution, we can create a quadratic equation

18. y = 80 – 2x A = (x)(y) 2x + y = 80 A = (x)(80 – 2x) = 80x – 2x 2 = -2x 2 + 80x Complete the square to convert, or…….

19. A = -2x 2 + 80x Find the change in Area with respect to the change in x. Make the derivative equal to zero and solve for x (to determine the max point)

20. Find y 2x + y = 80 2(20) + y = 80 y = 40 The width The length

21. 20m 40 m A = 20m X 40m = 800m 2 P = 12m + 56m + 12m = 80m

22. Consider a dance recital… At $3.00 a ticket, all 800 seats will be sold. For every $1.00 increase in ticket price, 100 fewer seats will be sold. What ticket price will maximize the profit?

23. Try a few calculations: R = (3)(800) = $2400.00 Increase by $1.00 R = (4)(700) = $2800.00 Increase by $7.00 R = (10)(100) = $1000.00

24. 1.Maximize the Revenue : R 2.Let x be the number of dollar increases in the ticket price. R = (cost per ticket)(number of tickets sold) = (3)(800) = (3 + x)(800 – 100x) = -100x 2 + 500x + 2400

25. Find the change in revenue with respect to the increase in ticket price. Equate to zero and solve for x 0 = -200x + 500 2.5 = xSub to determine the maximum Revenue $3025 = R

26. 3. The equation already has 1 variable 4. Complete the square to convert R = -100x 2 + 500x + 2400 = -100(x 2 – 5x) + 2400 = -100(x 2 – 5x +6.25 – 6.25) + 2400 = -100(x 2 – 5x + 6.25) + 625 + 2400 = -100(x – 2.5) 2 + 3025

27. sheet 1,2,4, 9,10 13,15 Pg 204 read example 1 Pg 205 read example 2 Pg 206 1,4,6,7,9,11