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Optimization: maximum curved surface area of a cylinder in sphere

  1. Jun 8, 2013 #1
    θ1. The problem statement, all variables and given/known data
    The attached diagram depicts a sphere with several variables: the height of the cylinder, the radius of the cylinder and an angle. All that has been given to me is the hypotenuse of a triangle used.


    2. Relevant equations
    To my knowledge I was told to use 2sinθcosθ=sin2θ


    3. The attempt at a solution
    Originally I attempted to get everything in terms of different variables and then substitute them into an equation to solve for one variable. Therefore, I used Pythagoras's Theorem and got:
    8^2=y^2+r^2. I used y^2 to denote the unknown height within the triangle.
    From here I used trigonometric functions and assumed the triangle was right angled. From here I got cosθ=y/8 and sinθ=r/8. And after this I tried substituting and it got very messy.

    I tried substituting for one variable, then re-arranging both trig equations to get one variable and then I attempted to eliminate it to no avail.

    Any help would be highly appreciate.
     

    Attached Files:

  2. jcsd
  3. Jun 8, 2013 #2

    tiny-tim

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    hi jackscholar! :smile:

    why are you using pythagoras? :confused:

    you have one variable, θ, and you need to find the surface area of the cylinder (is that just the side, or the top and bottom also?) as a function of θ

    start again …

    just use 8cosθ and 8sinθ :wink:
     
  4. Jun 8, 2013 #3
    I need to find the side, or at least thats what I think is meant by curved surface. I need to maximise the surface area of the curved rectangle and do so by finding the heigh and radius, or at least thats what is said. How do I go about doing it using 8cosθ and 8sinθ?
     
  5. Jun 8, 2013 #4

    tiny-tim

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    well, if you know θ (and 8), aren't the height and radius obvious? :confused:
     
  6. Jun 8, 2013 #5
    I don't know theta, it hasn't been given to me.
     
  7. Jun 8, 2013 #6

    tiny-tim

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    ah, that's because θ is your variable

    you write the surface area as a function of θ, and then you maximise it! :smile:
     
  8. Jun 8, 2013 #7
    So what now then? 8cosθ=radius and 8sinθ=height. That is what I have gathered. The surface area for the rectangle would be height multiplied by circumference. So the surface area would then be (8sinθ) (2*pi*8cos(θ))=curved rectangle. Then do I expand that, differentiate and find that will give me the maximum area?
     
  9. Jun 8, 2013 #8

    tiny-tim

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    yes :smile:

    or you could …
     
  10. Jun 8, 2013 #9
    Thank you very much :)
     
  11. Jun 8, 2013 #10
    So it then becomes 64∏sin(2θ). After that it is differentiated to become 128∏cos(2θ)=0. Therefore once re-arranged it becomes cos(2θ)=0, then apply the inverse of cos to both sides and divide by two...this gives ∏/4? Does this also apply to 3∏/4, 5∏/4, 7∏/4...(2n-1)∏/4 or something like that?
     
  12. Jun 8, 2013 #11

    tiny-tim

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    yes (except i think you're a factor of 4 out)
    yes, from differentiating, ∏/4 will be either a minimum or a maximum :smile:

    but couldn't you have worked that out just by looking at sin(2θ), ie without differentiating? :wink:

    (and without the worry of whether it's a minimum or a maximum!)
    yees …

    but what physical meaning would 3∏/4 etc have?? :confused:
     
  13. Jun 8, 2013 #12

    LCKurtz

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    I get that he's a factor of 2 off. Also, for what it's worth, that picture doesn't make any sense. It has the "vertical" height going from the center of the cylinder on the top through the center of the sphere to the circumference of the cylinder. That wouldn't be a straight vertical line.
     
  14. Jun 8, 2013 #13
    It was the best I could do given the diagram I had copied. Sorry for any inconvenience.
     
  15. Jun 9, 2013 #14

    tiny-tim

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    the accompanying description in the first post was clear, and the diagram helped by showing what the symbols meant …

    so the rest is just artistic licence! :smile:
     
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