## hawkfalcon 3 years ago The height of triangle ACE is 12 and base AE = 20. A rectangle is inscribed in triangle ACE as shown below.? a) Express the rectangle’s area as a function of x and y. Use the fact that the triangle ACE is similar to triangle BCD to write y as a function of x. c)Express the rectangle’s area as a function of only x. d) Find the maximum possible area of the rectangle.

1. hawkfalcon

|dw:1345995812545:dw|

2. hawkfalcon

Nobody knows what to do?:#

3. hawkfalcon

@Zekarias ?

4. Zekarias

a) (12-y)x b) 12(x-0.1x^2) c) 30 sq units

5. hawkfalcon

...How?

6. Zekarias

Is "How" for all of it?

7. hawkfalcon

Just b and c

8. hawkfalcon

especially b... :O

9. Zekarias

consider the perpendicular line form C, we can come out with the ratio 12/y =10/x, where 10 is half the rectangles side,20. Then simplify it so as to inset the result in to the equation in 'a'. To 'c' just use the application of derivative.

10. hawkfalcon

Okay. Makes sense I guess:p thanks!

11. ganeshie8

small catch

12. ganeshie8

triangle ACE is similar to triangle BCD => height1/height2 = base1/base2

13. ganeshie8

12/y = 20/x y = (3/5)x

14. ganeshie8

this would give you max area 60

15. hawkfalcon

o.O But y needs to be written as a function of x.

16. ganeshie8

y = (3/5)x

17. ganeshie8

its a function of x

18. ganeshie8

linear

19. hawkfalcon

Oh yeah :3. So why did you use the entire base and @Zekarias used 1/2?

20. ganeshie8

since both are similar triangles, "ratio of bases" equals "ratio of heights"

21. hawkfalcon

Okay, that makes sense. :3 Can you explain how you got 60 though?

22. ganeshie8

write the area equation in x and optimize

23. ganeshie8

for optimizing, use either of the below two : 1) its a parabola facing down so max value = f(-b/2a) or 2) use calculus

24. ganeshie8

this q from conics, or calculus ?

25. hawkfalcon

calculus. Going into calculus next year and they gave us some prep stuff*.*

26. ganeshie8

oh ok, so we have this so far : A = x(12-y) -------(1) y = (3/5)x --------(2)

27. hawkfalcon

Okay, true.

28. ganeshie8

hmm which one moves ?

29. ganeshie8

ohhk yeah as we go down its moving a lot it seems...

30. ganeshie8

substitute (2) in (1)

31. hawkfalcon

x(12-(3/5)x)

32. ganeshie8

$$A = x(12 - \frac{3}{5}x)$$ $$A = 12x- \frac{3}{5}x^2)$$

33. hawkfalcon

Okayy and solve for x

34. ganeshie8

nope we need to optimize A

35. ganeshie8

find maximum possible value of A, right ?

36. hawkfalcon

Oh.

37. hawkfalcon

Ya

38. ganeshie8

theres a process for finding maximum/minimum values of any funcitons

39. ganeshie8

you familiar with differentiation ?

40. hawkfalcon

... I guess I havenet learned that yet..

41. ganeshie8

its simple... if you knw differentiation already

42. hawkfalcon

I think we are learning that next year :( (monday)

43. ganeshie8

just if you knw derivative of x, and x^2, its enough

44. ganeshie8

oh ok then you cannot use calculus, lets switch to conics. you knw parabolas, right

45. hawkfalcon

Okay, I don't know how to do that.. Maybe I should make a new question asking for the dirivative of A=12x-3/4x^2

46. hawkfalcon

Yes.

47. hawkfalcon

You said f(-b/2a)

48. ganeshie8

yes

49. ganeshie8

A = 12x - (3/5) x^2

50. ganeshie8

$$A = \frac{-3}{5}x^2 + 12x$$

51. hawkfalcon

Yes

52. ganeshie8

a = -3/5 b = 12

53. hawkfalcon

Cool, thanks, just plug and chug now:D

54. ganeshie8

f(-b/2a) gives the vertex y co-ordinate, which is the maximum value

55. hawkfalcon

y=10

56. ganeshie8

|dw:1345999525420:dw|

57. ganeshie8

-b/2a = 10 f(-b/2a) = ?

58. ganeshie8

put x = -b/2a = 10, in area function

59. hawkfalcon

Okay.

60. hawkfalcon

60!:D

61. ganeshie8

if you remember, for any parabola, -b/2a is the x-coordinate of vertex f(-b/2a) is the y-coordinate of vertex

62. ganeshie8

great !!

63. hawkfalcon

Thank you so much! Now I can do the rest of these problems:D

64. ganeshie8

glad to hear ! calculus wud be much fun :) good luck !!