Chapter 3 – Solving Quadratics Word Problems Where You Have to Create the Equation

1. Chapter 3 – Solving Quadratics Word Problems Where You Have to Create the Equation

2. Agenda • Movement problems (creating the equation)* • Engineering problems (creating the equation)* • Revenue problems (creating the equation)* • Area problems (creating the equation)* • Integer problems (creating the equation) • Triangle problems (creating the equation) • * Indicates done within this lecture

3. Mental Health Break

4. Quadratic Word Problems Movement Creating the Equation – Ex. P.283, #14 • Angus is playing golf. The diagram (not to scale, shows him making a perfect shot to the hole. Determine the height of the ball when it is 15m from the hole by using the information in the diagram to determine a quadratic relation for height v. distance travelled

5. Quadratic Word Problems Movement Creating the Equation – Ex. P.283, #14

6. Quadratic Word Problems Movement Creating the Equation – Ex. P.283, #14

7. Quadratic Word Problems Movement Creating the Equation – Ex. P.283, #14

8. Quadratic Word Problems Movement Creating the Equation – Ex. P.283, #14 • To find the height we will need the equation • To find the equation we need the zeros and one other point • Zeros… • x of vertex… • y of vertex…

9. Quadratic Word Problems Movement Creating the Equation – Ex. P.283, #14 • To find the height we will need the equation • To find the equation we need the zeros and one other point • Zeros are 0 and 100 • x of vertex = (0+100)/2 = 50 • y of vertex… 10 (height of the tree in the middle)

10. Quadratic Word Problems Movement Creating the Equation – Ex. P.283, #14 • Use y = a(x – S)(x – t) • put in the zeros of 0 and 100 for S & t* • put in the point (50, 10) for x & y • *Note – you would end up with the same “a” if you used -50 and 50 for the zeros and made x = 0 the centre of the graph

11. Quadratic Word Problems Movement Creating the Equation – Ex. P.283, #14 • y = a(x – S)(x – t) • 10 = a(50 – 0)(50 – 100) 10 = a(50)(-50) 10 = a(-2500) 10/-2500 = a(-2500)/-2500 • a = -0.004 • y = -0.004x(x – 100) • y = -0.004x²+0.4x

12. Quadratic Word Problems Movement Creating the Equation – Ex. P.283, #14 • y = -0.004x²+0.4x • We shouldn’t cannot plug in an x of 15m because we are 15m from where the ball lands • If we did, we could argue symmetry however • Ideally, we plug in 85m for x because 100m – 15m = 85m and we can use that distance to then calculate the height of the ball at that time

13. Quadratic Word Problems Movement Creating the Equation – Ex. P.283, #14 • y = -0.004x² + 0.4x • If we plug in x of 85m, we can calculate y (the height of the ball at that time) • y = -0.004(85)² + 0.4(85) • y = -0.004(7225) + 34 • y = -28.9 + 34 • y = 5.1m • So, the height of the ball 15m from the hole is 5.1m

14. Quadratic Word Problems Movement Creating the Equation – Ex. P.283, #14

15. Quadratic Word Problems Engineering Creating the Equation – Ex. P.284, #16 • The second span of the Bluewater Bridge, in Sarnia, Ontario, is supported by a pair of steel parabolic arches. The arches are set in concrete foundations that are on opposite sides of the St. Clair River 281m apart. The top of each arch rises 71m above the river. Determine the algebraic expression that models the arch and find the height of the span 10 metres from the left side of the river.

16. Quadratic Word Problems Engineering Creating the Equation – Ex. P.284, #16

17. Quadratic Word Problems Engineering Creating the Equation – Ex. P.284, #16

18. Quadratic Word Problems Engineering Creating the Equation – Ex. P.284, #16 • To find the height we will need the equation • To find the equation we need the zeros and one other point • Zeros… • x of vertex… • y of vertex…

19. Quadratic Word Problems Engineering Creating the Equation – Ex. P.284, #16 • To find the height we will need the equation • To find the equation we need the zeros and one other point • Zeros are 0 and 281 • x of vertex = (0+281)/2 = 140.5 • y of vertex… 71 (height of the span of bridge in the middle)

20. Quadratic Word Problems Engineering Creating the Equation – Ex. P.284, #16 • Use y = a(x – S)(x – t) • put in the zeros of 0 and 281 for S & t* • put in the point (140.5, 71) for x & y • *Note – you would end up with the same “a” if you used –140.5 and 140.5 for the zeros and made x = 0 the centre of the graph

21. Quadratic Word Problems Engineering Creating the Equation – Ex. P.284, #16 • y = a(x – S)(x – t) • 71 = a(140.5 – 0)(140.5 – 281) 71 = a(140.5)(-140.5) 71 = a(-19740.25) 71/-19740.25 = a(71)/-19740.25 • a = -0.004 • y = -0.004(x - 0)(x – 281) • y = -0.004x(x – 281) • y = -0.004x² + 1.124x

22. Quadratic Word Problems Engineering Creating the Equation – Ex. P.284, #16 • y = -0.004x²+1.124x • We can plug in an x of 10m

23. Quadratic Word Problems Engineering Creating the Equation – Ex. P.284, #16 • y = -0.004x² + 1.124x • If we plug in x of 10m, we can calculate y (the height of the ball at that time) • y = -0.004(10)² + 1.124(10) • y = -0.004(100) + 11.24 • y = -0.4 + 11.24 • y = 10.84 • So, the height of the bridge span 10m from the left side of the river is 10.84m

24. Quadratic Word Problems Revenue Creating the Equation • Recall, R = price x # sold • Profit is the revenue of a company after expenses are subtracted • Breakeven takes place when profit (not revenue) is zero • Maximum profit or revenue takes place at the vertex of the graph

25. Quadratic Word Problems Revenue Creating the Equation • Icky Bum Ltd. research shows that a $0.50 increase in the price of a box of baby wipes results in 5 fewer boxes of baby wipes being sold. The usual price of $20 for a box of baby wipes results in a sale of 280 boxes • What should the Icky Bum do? Should they “stick” with the current price?

26. Quadratic Word Problems Revenue Creating the Equation • Equation set-up is always the same for this type of Revenue problem

27. Quadratic Word Problems Revenue Creating the Equation

28. Quadratic Word Problems Revenue Creating the Equation • R = (orig. price + ∆ in price) (orig. # sold + ∆ in number sold) • R = (20 + 0.50x)(280 – 5x)

29. Quadratic Word Problems Revenue Creating the Equation • R = (orig. price + ∆ in price) (orig. # sold + ∆ in number sold) • R = (20 + 0.50x)(280 – 5x) • Our strategy is going to be to set-up the equation, find the zeros, take the average of the zeros to find the x of the vertex, and substitute x back in and solve for the vertex R which will give us the maximum revenue

30. Quadratic Word Problems Revenue Creating the Equation • R = (20 + 0.50x)(280 – 5x) • 0 = (20 + 0.50x)(280 – 5x) • 20 + 0.50x = 0 and 280 – 5x = 0 (set each bracket = 0) • 0.50x = -20 and -5x = -280 • x = -40 and x = 56

31. Quadratic Word Problems Revenue Creating the Equation • R = (20 + 0.50x)(280 – 5x) • Zeros are -40 and 56 • xv = xzero 1 + xzero 2 --------------------------- 2 • xv = (-40 + 56) --------................ xv = 8 2

32. Quadratic Word Problems Revenue Creating the Equation • R = (20 + 0.50x)(280 – 5x) • xv = 8 • R = (20 + 0.50(8))(280 – 5(8)) • R = (24)(240) • So, maximum revenue is $5,760 (note – previously, it was $20 x 280 = $5,600)

33. Quadratic Word Problems Revenue Creating the Equation

34. Quadratic Word Problems Area Creating the Equation • When creating quadratic area equations, you are quite often working with either the area equation alone, or with the perimeter equation as well. • When working with maximum area created by enclosure of so much material, you will need both equations • When you are working with a maximum area of some type of border, you are usually working only with the area equation

35. Quadratic Word Problems Area Creating the Equation • A = l x w (area) • P = 2l + 2w (perimeter)

36. Quadratic Word Problems Area Creating the Equation • Ex. Say you want to find equation showing the maximum enclosure area given a perimeter of 20m

37. Quadratic Word Problems Area Creating the Equation

38. Quadratic Word Problems Area Creating the Equation • Ex. Say you want to find equation showing the maximum enclosure area given a perimeter of 20m • P = 2l + 2w • 20 = 2l + 2w (substitute 20 in for P) • 10 = l + w (divide both sides by 2) • l = 10 – w (isolate l)

39. Quadratic Word Problems Area Creating the Equation • Ex. Say you want to find equation showing the maximum enclosure area given a perimeter of 20m • Now, substitute l = 10 – w into A = l x w • A = (10 – w)(w) (factored form) • A = 10w - w² (standard form) • A = -w² + 10w ( standard form)

40. Quadratic Word Problems Area Creating the Equation

41. Quadratic Word Problems Area Creating the Equation • A = -w² + 10w • 0 = - w² + 10w • 0 = -w(w + 10) • -w = 0 and w + 10 = 0 • Zeros are 0 and -10 • wv= -5 (take average of zeros) • A = -(5)²+10(5) = 25 • So, a width of 5m results in the maximum area of 25m²

42. Quadratic Word Problems Area Creating the Equation • Mr. and Mrs. B want to install a rectangular swimming pool measuring 10m x 20m. They want to put a deck of uniform width around the pool. They have budgeted $1,920 to spend on the deck and know that construction and material costs are $30/metre

43. Quadratic Word Problems Area Creating the Equation • Here we are using the area equation only, but we have to consider the two areas – that of the pool & that of the deck surrounding it • Let’s let x represent the width of the uniform deck surrounding the pool • Then, let’s state what the “new” overall length and width are of the pool & the deck together

44. Quadratic Word Problems Area Creating the Equation

45. Quadratic Word Problems Area Creating the Equation

46. Quadratic Word Problems Area Creating the Equation • We know the inside area is: • A = 10 x 20 = 200m² • For the outside area we have to use other information given to determine it. We know that the B.’s have budgeted $1,920 to spend on the deck and that construction & material costs are $30 per metre. • How can we get an area out of this?

47. Quadratic Word Problems Area Creating the Equation • We know that the B.’s have budgeted $1,920 to spend on the deck and that construction & material costs are $30 per metre. • How can we get an area out of this? • Divide $1920 by $30 = 64 • So, the total area of the pool and the deck will be 264 m²

48. Quadratic Word Problems Area Creating the Equation • So, what is our equation? • A = l x w • 264 = (20 + 2x)(10 + 2x) • 264 = 200 + 60x + 4x² • 0 = 4x² + 60x + 200 – 264 • 0 = 4x² + 60x – 64 • 0 = 4(x² + 15x – 16) • 0 = 4(x + 16)(x – 1) • x = -16 or x = 1 (reject -16) so, the width of the deck should be 1m

49. Quadratic Word Problems Area Creating the Equation

50. Quadratic Word Problems Integers Creating the Problem • The key with integer word problems is the translation of words into algebra