## anonymous one year ago please help calculus: application of maxima and minima Find the volume of the largest right circular cylinder that can be cut from a circular cone of radius 6 cm and height 9 cm the V should be 48pi cm^3

1. ganeshie8

what have you tried so far ?

2. anonymous

Looks like an optimization problem

3. anonymous

i've tried to solve it but i don't know what to do and also how to insert the derivatives

4. anonymous

they have the same SA right?

5. ganeshie8

How so ? Maybe, as a start draw a nice sketch of cylinder inside a cone and label the given dimensions

6. ganeshie8

I think, this problem is equivalent to finding the volume of the largest cylinder that can be inscribed inside a cone..

7. anonymous

ohw now i get the figure

8. anonymous

|dw:1443926956736:dw| like this?

9. anonymous

|dw:1443926942036:dw|

10. anonymous

Yeah there you go lol.

11. ganeshie8

That looks good! label the given dimensions also..

12. ganeshie8

show radius 6 and height 9 in the diagram

13. anonymous

|dw:1443937323135:dw| volume of cylinder $V = \pi r^2 h$

14. anonymous

Naice.

15. anonymous

|dw:1443927181720:dw|

16. anonymous

$V=\pi \left( 6-\frac{ 2h }{ 3 } \right)^2 h$ diff. and find stationary values

17. anonymous

and then find second derivative and see where it is maxima

18. Zarkon

it is a constrained min/max problem. plug stationary points and boundary points into volume formula

19. Zarkon

pick the largest

20. Zarkon

$0\le h\le 9$

21. anonymous

should simplify it? i'm confused

22. anonymous

i don't know what to do with the volume part :(

23. anonymous

$\frac{ dV }{ dh }=\pi[h*2(6-\frac{ 2h }{ 3 })*\frac{ -2 }{ 3 }+1\left( 6-\frac{ 2h }{ 3 } \right)^2]$ simplify and put dV/dh =0 and find values of h

24. anonymous

okay

25. anonymous

$=\pi \left( 6-\frac{ 2h }{ 3 } \right)\left[ \frac{ -4h }{ 3 }+6-\frac{ 2h }{ 3 } \right]$

26. anonymous

remember 0<h<9

27. anonymous

$=\pi \left( 6-\frac{ 2h }{ 3 } \right)\left( 6-2h \right)$ i think now you can complete

28. anonymous

first bracket gives h=9,rejected as h<9

29. anonymous

only value is h=3 find second derivative and check for maxima

30. anonymous

then find r and finally V by above formula.

31. anonymous

can you solve further?

32. anonymous

may i ask how you get the factor on the dV/dh?

33. anonymous

you let out the common factor?

34. Zarkon

using the 2nd D is a waste of time...you don't need it

35. anonymous

$-\frac{ 4h }{ 3 }\left( 6-\frac{ 2h }{ 3 } \right)+\left( 6-\frac{ 2h }{ 3 } \right)^2$

36. anonymous

now take the common factor out

37. anonymous

you have to show it is maximum or minimum value. Here by chance you get only one value,you may get more values.

38. Zarkon

you need to look at h=0,3,9 (end poins and critical values) V(0)=V(9)=0 V(3) will be the absolute max since it is positive

39. anonymous

when h=0 ,V=0 when h=9,r=0 V=0 actually cylinder with zero radius or zero height so no cylinder with these dimensions that is why 0<h<9

40. Zarkon

you can use 0 and 9...you get a degenerate cylinder

41. Zarkon

the math works out. you don't need the 2nd D

42. anonymous

H=3 Then r=4 V=48pi

43. anonymous

I got it thank you very much