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Feb 22. Exercise 6

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Transcript of "Feb 22. Exercise 6"

1. 1. Review of Exercise 6: Transformations of Quadratic Functions 3
2. 2. 1. A farmer wishes to build a rectangular pen along one side of his barn. If he has 80 metres of fencing, find the dimensions that will yield a maximum area. w 2w + L = 80 l barn L = 80 - 2w Area = L * w w
3. 3. w L = 80 - 2w l barn Area = L * w w A = (80 - 2w) * w 2 A = 80w - 2w
4. 4. 2 A = 80w - 2w 2 A = -2{w - 40w} 2 A = -2{w - 40w + 400 - 400} 2 A = -2{(w-20) - 400} 2 A = -2(w-20) + 800
5. 5. w L = 80 - 2w l barn Area = L * w w 2 A = -2(w-20) + 800 Area is a maximum when w = 20. 2 The maximum area is 800 m 800 = L * 20 L = 40 m
6. 6. 2. Find 2 positive numbers whose sum is 13 if the sum of their squares is a minimum 2 positive numbers: a, b a + b = 13 2 2 a +b =y where y is a minimum
7. 7. 2 2 a +b =y where y is a minimum a + b = 13 a = 13 - b 2 2 a +b =y 2 2 (13 - b) + b = y
8. 8. 2 2 (13 - b) + b = y 2 2 (169 - 26b + b ) + b = y 2 y = 2b - 26b + 169 2 y = 2{b - 13b} + 169
9. 9. 2 y = 2{b - 13b} + 169 2 2 2 y = 2{b - 13b + 6.5 - 6.5 } + 169 2 2 y = 2{(b - 6.5) - 6.5 } + 169 2 2 y = 2(b - 6.5) - 2(6.5 ) + 169 2 y = 2(b - 6.5) - 84.5 + 169 2 y = 2(b - 6.5) + 84.5
10. 10. a + b = 13 2 2 a +b =y where y is a minimum 2 y = 2(b - 6.5) + 84.5 y is a minimum when b = 6.5 a + 6.5 = 13 a = 13 - 6.5 a = 6.5
11. 11. 3. A projectile is shot straight up from a height of 6 m with an initial velocity of80 m/s. Its height in meters above the ground after t seconds is given by the equation 2 h = 6 + 80t - 5t . After how many seconds does the projectile reach its max height, and what is this height? max height 6m
12. 12. 2 h = 6 + 80t - 5t 2 h = - 5t + 80t + 6 2 h = -5{t - 16t} +6 2 h = -5{t - 16t + 64 - 64} +6 2 h = -5{(t - 8) - 64} + 6 2 h = -5(t - 8) - (-5)(64) +6
13. 13. 2 h = -5(t - 8) - (-5)(64) +6 2 h = -5(t - 8) + 326 The maximum height is reached after 8 seconds. The maximum height is 326 metres.
14. 14. 4. A survey found that 400 people will attend a theatre when the admission price is 80 cents. The attendance decreases by 40 people for each 10 cents added to the price. What price admission will yield the greatest receipt? Profit = Tickets * Cost x = number of times the ticket price is increased
15. 15. Profit = Tickets * Cost x = number of times the ticket price is increased T = 400 - 40x C = .8 + .1x P = (400 - 40x) (.8 + .1x)
16. 16. P = (400 - 40x) (.8 + .1x) 2 P = 320 + 40x - 32x - 4x 2 P = -4x + 8x + 320
17. 17. 2 P = -4x + 8x + 320 2 P = -4{x - 2x} + 320 P = -4{x 2 - 2x + 1 - 1} + 320 P = -4{(x - 1) 2 - 1} + 320 2 -1(-4) + 320 P = -4(x - 1) 2 + 324 P = -4(x - 1)
18. 18. P = -4(x - 1) 2 + 324 x = number of times the ticket price is increased Profit = Tickets * Cost Profit will be a maximum when x = 1. C = .8 + .1x Cost of each ticket will give the maximum profit when C = .8 + .1(1) C = \$0.90
19. 19. 5. Find 2 positive numbers whose sum is 13 and whose product is a maximum. 2 positive numbers: a, b a + b = 13 a*b=c where c is a maximum
20. 20. a*b=c where c is a maximum a + b = 13 a = 13 - b a*b=c (13 - b) * b = c 13b - b 2=c
21. 21. 13b - b2=c c = -b2 + 13b c = -1{b2 - 13b} c = -1{b2 - 13b + 6.52 - 6.52} 2 - 6.52} c = -1{(b - 6.5)
22. 22. c = -1{(b - 6.5) 2 - 6.52} c = -1(b - 6.5) 2 - (-1)6.52 c = -1(b - 6.5) 2 + 42.25
23. 23. c = -1(b - 6.5) 2 + 42.25 c is a maximum when b = 6.5. a + b = 13 a + 6.5 = 13 a = 6.5
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