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Optimization Problem Maximum Volume

  1. Jul 24, 2012 #1
    1. The problem statement, all variables and given/known data
    a cylinder can be inscribed upright in a circular cone with radius 4 and height 7. What is the maximum volume of a cylinder that can be inscribed inside of the cone.


    2. Relevant equations
    Image of the problem

    http://s17.postimage.org/67akeibxb/Cylinder.png

    3. The attempt at a solution

    let rc = radius of the cone Let rt=radius of cylinder

    Let hc=height of cone let ht= height of cyldiner

    ht/hc = (4-rt)/rc

    ht/7=(4-rt)/4
    ht=7-7/4(rt)

    D: 0<r<4

    Vt=Volume of cylinder
    Vt=Pi(rt)^2(ht)
    vt=pi(rt)^2(7-7/4(rt))

    vt=7pi(rt)^2 - 7/4(pi)(rt)^3

    vt'=14pi(rt) - 21/4(pi)(rt)^2

    vt'=0

    0 =14pi(rt) - 21/4(pi)(rt)^2
    0=7pi(rt)[2-3/4rt]

    r=0, not true
    0 = 2-3/4rt
    -2 = -3/4rt
    rt=8/3=2.67


    v'(2.5)= + numbrer
    v'(2.7)= - number

    therefore there is a maximum value

    there fore rt=2.67 is the value which will acheive maximum volume.

    h=7-7/4rt
    h=2.33

    vt=Pi(rt)^2h
    vt=52.18m^3

    Therefore maximum volume of a cylinder that can be inscribed in a cone is 52.18 m^2


    Okay i think i have this correct, i'm just worried i messed up my numbers and i was wondering if some one could help me out.. i've doouble and triple checked it myself... Also can anyone tell me if i'm missing any information that would be valuable?>
     
    Last edited: Jul 24, 2012
  2. jcsd
  3. Jul 24, 2012 #2

    LCKurtz

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    Sure it can. So what? Is there a question here?

    Solution to what?

    Your work for what I suppose the problem was looks OK. But if you are going to hand it in for grading you best explain why you know the end points r = 0 and r = 4 don't give the maximum. Also there is no reason to use decimal approximations anywhere in this problem. Use exact fractions.
     
  4. Jul 24, 2012 #3
    Ugh, i cant believe i forgot to put what it was asking for, i re edited it in the original post.

    And I believe the reason they cannot be a maximum is the radius cannot be zero, because then it wouldn't be a cylinder, and the radius can't be 4 because it cannot be greater or equal to the radius of the cone¿
     
  5. Jul 25, 2012 #4

    HallsofIvy

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    Well, in the cases r= 0 and r= 4, we could think of the figure as a "degenerate" cylinder. I would NOT say 'the radius cannoot be greater than or equal to the radius of the cone." Clearly it cannot be greater because the cylnder must be in the cone but you would need to show that it cannot be equal. And, again, it can be if you accept "degenerate" cylinders with volume 0. More to the point here, 0 cannot be the MAXIMUM volume because it is easy to see that there are cylinders of positive volume in the cone.
     
  6. Jul 25, 2012 #5
    Why is simply not rationalizing that the cylinder would not fit in the cone properly if it's radius was greater than that of the cone, and How can this be shown?


    So in this case can i say it cannot be <0 because the degenerate case is not accepeted therefore we must have a cone with positive volume to have a max volume?
     
  7. Jul 25, 2012 #6

    HallsofIvy

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    I said nothing about '< 0' nor did I say that degenerate cases could not be accepted. I only said that 0 cannot be a maximum because it is smaller than the positive values.
     
  8. Jul 25, 2012 #7
    Okay, and in regard to the cone's radius of 4, what is the way to prove this in math?
    Here is my idea...

    v(4)= 0

    Therefore since the volume is 0, and all positive volumes are greater than this.

    If you take any number greater than r=4 your volume will be negative(Ie:v(4.1), and this does not comply with what you said earlier about the fact that the cylinder has (edit)***POSITIVE VOLUMES IN THE CONE**** in this case.

    So therefore the the radius must be between 0<r<4
     
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