## anonymous 5 years ago A tough one. Find the largest possible area for A(theta)=49sin(2theta)

1. anonymous

$A(\Theta)=49\sin 2(\Theta)$

2. amistre64

you know how to do derivatives?

3. anonymous

haha some but not with trigonometric functions. I am in precalc

4. amistre64

ok...lets try this route; the largest area of any "sin" is a full circle right?

5. anonymous

yes, 2pi

6. amistre64

then we need to find a circle that has the same "information" is the given equation.

7. amistre64

Do you know the area of a sector? when it involves 2 sides and the sin of an angle?

8. anonymous

not yet...I think thats in an upcoming chapter.

9. amistre64

ok... and maybe I saw this wrong to begin with...consider this: the area of a triangle is: (1/2) (base) (height) right?

10. anonymous

yes

11. amistre64

How do we find the height of a triangle when we know an angle? does sin(t) = y/r?

12. anonymous

yes

13. amistre64

like this

14. anonymous

yes...I understand that

15. amistre64

good, then we need to determine the value of "t" that will give us the largest area for the triangle....or thats how I understand this problem

16. anonymous

and the biggest sin you can get is 1

17. anonymous

would it help to tell you that I have a rectangle inscribed inside a semicircle with a radius of 7cm.

18. anonymous

and that is the equation for the area

19. amistre64

i keep seeing this 2 different ways...for example: the area of any given sector of a circle is define as (theta)(r^2)/2

20. amistre64

it would..... all the information helps :)

21. anonymous

22. amistre64

like this?

23. amistre64

like this?

24. anonymous

exactly!

25. amistre64

yeah....thatd a been helpful to know lol

26. anonymous

haha...wow...sorry

27. amistre64

which angle is the theta...near the center of up near a corner?

28. anonymous

near the center

29. amistre64

if I recall correctly, the angle needs to be 45 degrees. so 2t = 45 t = 22.5

30. amistre64

the largest area you can get is formed by a square.... and you need 2 squares side by side to make this the max area under the half circle

31. anonymous

so would my dimension them be 7 by 14?

32. amistre64

2 sides equal 7sin(45); and the other 2 sides are 28sin(45) 4.95 is one side... 9.90 is the other side...... it gives you an area about equal to 49.005

33. anonymous

how did you get these?

34. amistre64

magic :) first I killed a live chicken; then a spread its blood around the bedposts......

35. anonymous

haha seriosuly!...thats how I feel about these problems! You need to do voodoo to get an answer!!

36. amistre64

we take the equation for a half circle: y = sqrt(7^2-x^2) and the Area of the "rectangle" to be maximized. A = xy ; substitute y = sqrt(49 - x^2) into this equation

37. amistre64

A = x(sqrt(49-x^2)) now find the derivative.... A' = x(-x/(sqrt(49-x^2))) +1(sqrt(49-x^2)) and make that equal to zero.... then solve for x :)

38. anonymous

thanks!

39. amistre64

its calculus....but it works :)

40. amistre64

the other way is just to notice that the largest area that can be produced in a quarter of a circle is a square....a square has a 45 degree angle where we need it down there

41. anonymous

gotcha!!

42. anonymous

thanks! have a good night!

43. amistre64

youre welcome :) Gnite