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Math Help - Maximum volume of tower

  1. #1
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    Maximum volume of tower

    1) The bottom part of the tower is cylinder-shaped and the upper part of the tower is cone-shaped. Equilateral triangel is for the axis segment to the cone-shaped part.
    How much must be the radius of the cylinder-shaped bottom part that the volume of the tower would be maximum if the tower axis segment perimeter is 90 meters ?

    2) What are generally the conditions that define the maximum or minimum volume and how to use them ?
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  2. #2
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    Hard to understand.

    "Equilateral triangel is for the axis segment to the cone-shaped part"
    I understand that means if 2R is the diameter of the cone, then 2R is also the length of the slanting side of the cone.

    How about
    "...the tower axis segment perimeter is 90 meters" ?

    I assume the radius, R, of the cylinder is the same as the radius of the cone.
    If the height of the cylinder is h, does it mean that
    6R +2h = 90 ? ------silhouette or outline of the whole tower.
    Or, 4R +2h = 90 ? -------silhouette of the cylinder only.

    ------------------
    I will come back after about 14 hours from now. Got to go to work.
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  3. #3
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    "Equilateral triangel is for the axis segment to the cone-shaped part"
    I though out that the side of the equilateral triangel is equal to the diameter of the cone and equal to the slanting side of the cone as well.

    "...the tower axis segment perimeter is 90 meters" ?
    The radius, R, of the cylinder is the same as the radius of the cone.

    But there is one thing that is confusing me. The tower axis segment consist of equilateral triangel and square or rectangle.
    I think something like that: "6R +2h = 90 meters --- outline of the whole tower."
    What about 6R ? Why 6R ? If the height of the cylinder is equal to equilateral triangel then the height of the tower is really 2h.
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  4. #4
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    What is hard to uinderstand is your interpretation of the original problem.

    "axis segment to the cone-shaped part."
    "tower axis segment perimeter"

    Could you say another English term for axis segment?
    Is that cross section?, vertical section?

    Or, I thought you might show a picture?

    The problem, or solution, is easy. Show me the figure and I will show you a solution.
    Or, if there is an answer in the source of the question/problem, then it should be:
    R = 12.383 m for max volume of the whole tower.
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  5. #5
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    "axis segment to the cone-shaped part."
    The segment of the axle.
    "tower axis segment perimeter"
    This is the perimeter of the triangle and square. The triangle and square are into the tower. The triangle is into the cone, and the square is into the cylinder. Both are in the middle of the cone and cylinder.

    And the perimeter of the green line is 90 meters.

    Maybe "perimeter" is not the right word. I used my dictionary and found the word GIRTH - Distance round something.
    Attached Thumbnails Attached Thumbnails Maximum volume of tower-mata.jpg  
    Last edited by totalnewbie; July 29th 2005 at 04:19 AM.
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  6. #6
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    Thanks for the figure. It is really what I thought it should be.
    Like I said, for max volume of the whole tower, the radius "r" of the cylinder, and of the cone, is 12.383 meters.

    Cone:
    Radius = r
    Diameter = 2r
    Slant height = 2r also.
    Vertical height = sqrt[(2r)^2 -(r)^2] = sqrt[4r^2 -r^2] = r*sqrt(3)
    Volume = (1/3)pi(r^2)(height) = (1/3)pi(r^2)[r*sqrt(3)] = [(sqrt(3))/3]pi(r^3) = (0.57735)pi(r^3)

    Cylinder:
    Radius = r also
    Height = h
    Volume = pi(r^2)h

    So, total volume of tower, V = (0.57735)pi(r^3) +pi(r^2)h --------(i)

    We find a way to express the "h" in terms of "r" so that we will have one independent variable only---for easy differentiation.

    Given:
    (starting from the leftmost of the bottom base of the cylinder, going up and clockwise) h +2r +2r +h +2r = 90
    6r +2h = 90
    3r +h = 45
    h = 45 -3r ----***

    Substitute that into (i),
    V = (0.57735)pi(r^3) +pi(r^2)(45 -3r)
    V = (0.57735)pi(r^3) +45pi(r^2) -3pi(r^3)
    V = 45pi(r^2) -(2.42265)pi(r^3) ----------(ii)

    V is max or min when dV/dr, or V', is zero.

    V' = 45pi[2r] -(2.42265 pi)[3r^2]
    V' = 90pi(r) -(7.26795 pi)(r^2)
    Set that to zero,
    0 = 90pi(r) -(7.26795 pi)(r^2)
    Divide both sides by (7.26795 pi r),
    0 = 12.383134 -r
    r = 12.383 m ------for max or min V.

    No need to test if that is for max or min V, because min V is zero here.

    Therefore, r = 12.383 m for max V ---------answer.

    ------------
    "2) What are generally the conditions that define the maximum or minimum volume and how to use them ?"

    That beats me.
    I don't know how to answer that.
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  7. #7
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    I took my pencil and tried to realize what you did.
    I wonder that you took the derivative from 'r'. Then r' into the volume formula means maximum volume, this formula: V' = 90pi(r) -(7.26795 pi)(r^2).

    Full answer is: -90/[sqrt(3)-9]
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  8. #8
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    Full answer is: -90/[sqrt(3)-9].
    Yes, and that is 12.383 if you simplify it.

    That was actually what I got even before you showed the drawing. I thought a "simple" r would be better than that "r = -90/[sqrt(3)-9]."

    Volume of cone = [(sqrt(3))/3]pi(r^3) -----let us use this then.

    Volume of cylinder = pi(r^2)h = pi(r^2)(45 -3r) = 45pi(r^2) -3pi(r^3)

    So,
    V = [(sqrt(3))/3]pi(r^3) +45pi(r^2) -3pi(r^3)
    V = [(sqrt(3))/3 -3]pi(r^3) +45pi(r^2)
    V' = [(sqrt(3))/3 -3]pi(3r^2) +45pi(2r)
    V' = [sqrt(3) -9]pi(r^2) +90pi(r)
    Set that to zero,
    0 = [sqrt(3) -9]pi(r^2) +90pi(r)
    Divide both sides by pi(r),
    0 = [sqrt(3) -9]r +90
    -90 = [sqrt(3) -9]r
    r = -90 / [sqrt(3) -9] ---------see?

    ---------------------------------------
    "I wonder that you took the derivative from 'r'. Then r' into the volume formula means maximum volume,...."

    Do you mean why I took derivative with respect to "r"?
    V = [(sqrt(3))/3 -3]pi(r^3) +45pi(r^2)
    To get the first derivative of V, where would I take if from? Where would I base it from? Not from "r"?

    r' ?
    What r' ?
    There is no r'.

    Do you mean r' = dr/dV?
    You mean we should have taken the derivative of "r" with respect to V?
    That is funny, totalnewbie.

    ----------
    Do I understand you don't know the reason yet why V' = 0 will lead to maximum V?

    Sorry, sometimes I don't understand your English, so I don't know what you really mean.
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  9. #9
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    Full answer is: -90/[sqrt(3)-9].
    Yes, and that is 12.383 if you simplify it.
    ----------------------------------------
    Yes, I knew that these are the same answers. Just exact answer is better. I was alway told that do not write an approximate answer. So, in my opinion "simple" r isn't the best answer. Ok, if you are used to write so, then let it be. But I am used to write exact answers.


    "Do I understand you don't know the reason yet why V' = 0 will lead to maximum V?"
    Yes, I didn't know that V' = 0 will lead to maximum V. I see it for the first time

    I don't understand that equation: dr/dV

    You got the first derivative of V. You had the formula: V = [(sqrt(3))/3 -3]pi(r^3) +45pi(r^2). And then the first derivative of r^3 is 3r^2 and the first derivative of r^2 is 2r, and you substituted these and then you got the formula V' = [(sqrt(3))/3 -3]pi(3r^2) +45pi(2r) = [sqrt(3) -9]pi(r^2) +90pi(r). I didn't get that point immediatley. I haven't seen that kind of way before.
    Last edited by totalnewbie; July 29th 2005 at 12:57 PM.
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  10. #10
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    dr/dV, by itself, is derivative of r with respect to V.
    dr/dV should be your r'.

    dr/dV means r is dependent on the change in V.

    -----------
    "I haven't seen that kind of way before."

    I see.
    Math has many ways to only one answer. You will learn that as you go on.

    And, for me, there is no best answer/solution. Answers are answers (or solutions leading to the answers), whether they are simple, good, elegant, backwards, by brute force, by trial and error, best, .....

    In some Math Forums, the arguments are mostly on elegant versus backwards solutions. The Mathheads (people with degrees in Higher Math) tend to look down on solutions that are not "elegant".
    When I have a chance, I __________ these Mathheads.
    Last edited by ticbol; July 29th 2005 at 06:10 PM.
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  11. #11
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    Quote Originally Posted by totalnewbie
    2) What are generally the conditions that define the maximum or minimum volume and how to use them ?
    I have analysed many excercises whose task is to find maximum or minimum volume(area) and I can answer my quotation now. All these excercises based on the extremum. The general condition is simple. If the second derivative of a function is smaller than 0, then it is a maximum volume(area). If bigger, then it is a minimum volume(area).
    And according to my theme it is easy to show it. The second derivative of the function respect to r is V''=[sqrt(3)-9]*pi*2r+90*pi. Substituting radius r into the formula of the second derivative we get the answer -5890.37 which is smaller than 0 and it means the maximum volume.
    And according to my theme I can also say that the minimum area is when radius is 0. Substituting 0 into the formula of the second derivative we get the answer 90*pi which is bigger than 0 and it means the minimum volume.
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