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jarp0120

  • 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

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  1. ganeshie8
    • one year ago
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    what have you tried so far ?

  2. Jhannybean
    • one year ago
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    Looks like an optimization problem

  3. Jarp0120
    • one year ago
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    i've tried to solve it but i don't know what to do and also how to insert the derivatives

  4. Jarp0120
    • one year ago
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    they have the same SA right?

  5. ganeshie8
    • one year ago
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    How so ? Maybe, as a start draw a nice sketch of cylinder inside a cone and label the given dimensions

  6. ganeshie8
    • one year ago
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    I think, this problem is equivalent to finding the volume of the largest cylinder that can be inscribed inside a cone..

  7. Jarp0120
    • one year ago
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    ohw now i get the figure

  8. Jarp0120
    • one year ago
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    |dw:1443926956736:dw| like this?

  9. Jhannybean
    • one year ago
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    |dw:1443926942036:dw|

  10. Jhannybean
    • one year ago
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    Yeah there you go lol.

  11. ganeshie8
    • one year ago
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    That looks good! label the given dimensions also..

  12. ganeshie8
    • one year ago
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    show radius 6 and height 9 in the diagram

  13. anonymous
    • one year ago
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    |dw:1443937323135:dw| volume of cylinder \[V = \pi r^2 h\]

  14. Jhannybean
    • one year ago
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    Naice.

  15. Jarp0120
    • one year ago
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    |dw:1443927181720:dw|

  16. anonymous
    • one year ago
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    \[V=\pi \left( 6-\frac{ 2h }{ 3 } \right)^2 h\] diff. and find stationary values

  17. anonymous
    • one year ago
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    and then find second derivative and see where it is maxima

  18. Zarkon
    • one year ago
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    it is a constrained min/max problem. plug stationary points and boundary points into volume formula

  19. Zarkon
    • one year ago
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    pick the largest

  20. Zarkon
    • one year ago
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    \[0\le h\le 9\]

  21. Jarp0120
    • one year ago
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    should simplify it? i'm confused

  22. Jarp0120
    • one year ago
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    i don't know what to do with the volume part :(

  23. anonymous
    • one year ago
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    \[\frac{ dV }{ dh }=\pi[h*2(6-\frac{ 2h }{ 3 })*\frac{ -2 }{ 3 }+1\left( 6-\frac{ 2h }{ 3 } \right)^2]\] simplify and put dV/dh =0 and find values of h

  24. Jarp0120
    • one year ago
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    okay

  25. anonymous
    • one year ago
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    \[=\pi \left( 6-\frac{ 2h }{ 3 } \right)\left[ \frac{ -4h }{ 3 }+6-\frac{ 2h }{ 3 } \right]\]

  26. anonymous
    • one year ago
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    remember 0<h<9

  27. anonymous
    • one year ago
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    \[=\pi \left( 6-\frac{ 2h }{ 3 } \right)\left( 6-2h \right)\] i think now you can complete

  28. anonymous
    • one year ago
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    first bracket gives h=9,rejected as h<9

  29. anonymous
    • one year ago
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    only value is h=3 find second derivative and check for maxima

  30. anonymous
    • one year ago
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    then find r and finally V by above formula.

  31. anonymous
    • one year ago
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    can you solve further?

  32. Jarp0120
    • one year ago
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    may i ask how you get the factor on the dV/dh?

  33. Jarp0120
    • one year ago
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    you let out the common factor?

  34. Zarkon
    • one year ago
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    using the 2nd D is a waste of time...you don't need it

  35. anonymous
    • one year ago
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    \[-\frac{ 4h }{ 3 }\left( 6-\frac{ 2h }{ 3 } \right)+\left( 6-\frac{ 2h }{ 3 } \right)^2\]

  36. anonymous
    • one year ago
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    now take the common factor out

  37. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    you can use 0 and 9...you get a degenerate cylinder

  41. Zarkon
    • one year ago
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    the math works out. you don't need the 2nd D

  42. Jarp0120
    • one year ago
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    H=3 Then r=4 V=48pi

  43. Jarp0120
    • one year ago
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    I got it thank you very much

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