## anonymous 5 years ago please help me out with this one- a right circular cone has a surface area of 718 square inches. what dimensions (radius and height) will result in a maximum volume?

1. anonymous

I began the problem, and now it after 30 mins of work it is a nightmare.

2. Mertsj

What class is this for--calculus?

3. anonymous

yup

4. anonymous

so far, i've taken the SA eqn and tried to optimize it with substitution, but then after finding a derivative it just is bad.

5. Mertsj

6. amistre64

this one, jwg lol

7. Mertsj

yep

8. amistre64

is there a formula for the surface area of a rt circular cone?

9. Mertsj

SA=pir^2+pirl

10. amistre64

and perferably one for volume as well to relate the 2?

11. amistre64

1/3 something is all i can remember

12. amistre64

1/3 volume of a cylinder perhaps?

13. Mertsj

V=-1/3pir^2h

14. Mertsj

Exactly

15. amistre64

ok; looks like we are given an value for SA so that we can find an r perhaps?

16. Mertsj

The problem is the slant height adds a variable.

17. amistre64

i take it thats the "l" part in the SA

18. Mertsj

yes

19. amistre64

|dw:1327287244737:dw|

20. Mertsj

precisely

21. amistre64

l = sqrt(h^2 + r^2) then 718 = pi r^2 + pi r(sqrt(h^2+r^2)) 718/pir = r + (sqrt(h^2+r^2)) (718/pir) - r = sqrt(h^2+r^2) [(718/pir) - r]^2 = h^2+r^2 [(718/pir) - r]^2 - r^2 = h^2 i wonder of thats going according to plan :)

22. amistre64

at any rate we can define h in terms of r; aint pretty but doable

23. Mertsj

That should be progress. Wonder where the asker is.

24. amistre64

fetal position, sobbing maybe lol

25. amistre64

V = 1/3 pi r^2 h $V=\frac{pi\ r^2\ \sqrt{(\frac{718}{pi\ r}-r)^2-r^2}}{3}$ looks about right?

26. Mertsj

yes

27. Mertsj

Now I suppose we have to differentiate that and set to 0?

28. amistre64

the 1/3 pi is a constant that can be put to the side and we are left with deriving:$V=r^2\ ((\frac{718}{pi\ r}-r)^2-r^2)^{1/2}$

29. amistre64

yep, find the zeros and the undefineds to test

30. amistre64

$V'=r'^2\ ((\frac{718}{pi\ r}-r)^2-r^2)^{1/2}+r^2\ ((\frac{718}{pi\ r}-r)^2-r^2)'^{1/2}$ $V'=2r\ ((\frac{718}{pi\ r}-r)^2-r^2)^{1/2}+\frac{1}{2}r^2\ ((\frac{718}{pi\ r}-r)^2-r^2)^{-1/2}*((\frac{718}{pi\ r}-r)^2-r^2)'$ $V'=2r\ ((\frac{718}{pi\ r}-r)^2-r^2)^{1/2}+\frac{1}{2}r^2\ ((\frac{718}{pi\ r}-r)^2-r^2)^{-1/2}*(\frac{718}{pi\ r}-r)'^2-r'^2$ $V'=2r\ ((\frac{718}{pi\ r}-r)^2-r^2)^{1/2}+\frac{1}{2}r^2\ ((\frac{718}{pi\ r}-r)^2-r^2)^{-1/2}*2(\frac{718}{pi\ r}-r)(\frac{718}{pi\ r}-r)'-2r$ yep, its a doozie; the wolf might be quicker

31. Mertsj

That's what I did.

32. amistre64

..... slacker!! lol

33. Mertsj

proudly so

34. Mertsj

r=7.56

35. amistre64

yeah, i was gonna say around 8

36. amistre64

plug that value in to get the V max

37. Mertsj

I just ignored the miserable denominator and set the numerator to 0.

38. Mertsj

Must first find the height

39. amistre64

[(718/pir) - r]^2 - r^2 = h^2 plug in yer r

40. Mertsj

yep

41. amistre64

i dunno if solving in terms of h would have been easier

42. Mertsj

tooooooooooooooooooooooooooooo late

43. amistre64

lol, since h was buried i assume not

44. Mertsj

h= 21.37

45. amistre64

that should do it then

46. Mertsj

Except for checking to make sure those values result in the correct surface area.

47. amistre64

it has to, we did everything correctly

48. Mertsj

Of course...no mistakes...ever!!

49. amistre64

they wanted max volume

50. amistre64

the wolf helped so we can always pass the blame ;)

51. Mertsj

working on that.

52. Mertsj

1279.02= max volume

53. Mertsj

Good to have a scapegoat.

54. Mertsj

Hey!! It's right. SA = 717.9758

55. Mertsj

Thank you very much.