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anonymous

  • 5 years ago

Does anyone know any tips to follow for optimization problems and how to deal with the different shapes in them?

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  1. amistre64
    • 5 years ago
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    whats the problem? we dont all see what you see :)

  2. anonymous
    • 5 years ago
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    Find the height of the largest cylinder that can be placed inside a sphere whose radius is 4*the root of 3. I have the solution in front of me, but I don't understand how in a cylinder, the height will be divided into two, but in other shapes, a triangle inside a circle for example, that doesn't happen?

  3. amistre64
    • 5 years ago
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    whose radius? cyl or sphere?

  4. anonymous
    • 5 years ago
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    sphere

  5. amistre64
    • 5 years ago
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    we can take the cross section and just figure out the rectangle in a circle... right?

  6. anonymous
    • 5 years ago
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    well, what they've done is create a mini triangle inside the cylinder, using the radius one of the triangles sides and having (h) height, and (x) being the other sides and then we do pythagorean theorem

  7. amistre64
    • 5 years ago
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    like this right?

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  8. amistre64
    • 5 years ago
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    do we want to maximize the area of the cyl? or just the height?

  9. anonymous
    • 5 years ago
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    Like this: sorry for the cheesy sketch =P

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  10. amistre64
    • 5 years ago
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    area...volume; same diff :)

  11. amistre64
    • 5 years ago
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    its a good pic :) but without any restrictions to the cylindar; the max height can be close to the "diameter" of the sphere right? the question seens to be missing something

  12. anonymous
    • 5 years ago
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    In another problem, where there's a triangle inside a sphere, the height doesn't get divided like that, instead they take the whole height, and the perpendicular side would become (h-(whatever radius they give). How do I deal with all these different shapes?

  13. amistre64
    • 5 years ago
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    this is usually associated with the question: What is the largest volume right sylindar that can fit inside a sphere of radius (#)

  14. amistre64
    • 5 years ago
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    the shapes I could deal with; its the missing information to make the question plausible thats getting me :)

  15. anonymous
    • 5 years ago
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    yea, the solution in front of me seems pretty simple, but when i try to solve myself I always end up messing up somewhere...

  16. amistre64
    • 5 years ago
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    What does: "Find the height of the largest cylinder" mean?? I assume its talking about volume....but i cant tell :) the height of the cylindar is only limited by the diameter of the sphere....whats the answer they give?

  17. anonymous
    • 5 years ago
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    there is no missing info, from the sketch you get a constraint that's like this (r^2=48-h^2) and you use that for r^2 in your function which is the volume of the cylinder (V= pie*r^2h)

  18. amistre64
    • 5 years ago
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    So it IS volume we want to maximize....right?

  19. anonymous
    • 5 years ago
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    then you take the first derivative and let it equal zero, and solve for height. I guess I just thought there was a general rule that could be followed.

  20. amistre64
    • 5 years ago
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    which of these is the radius of the sphere? \[\sqrt[4]{3} \leftarrow \rightarrow 4\sqrt{3}\]

  21. anonymous
    • 5 years ago
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    second one

  22. amistre64
    • 5 years ago
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    good :) then we can work this easier :)

  23. anonymous
    • 5 years ago
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    the end reply for h=4 cm, since h was divided into two, the height would be 2h, therefore 8 cm =)

  24. amistre64
    • 5 years ago
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    it pretty much boils down to this; what is the largest area rectangle you can fit under a quarter top circle..

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  25. amistre64
    • 5 years ago
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    the equation of a circle is: x^2 + y^2 = r^2 y^2 = r^2 - x^2 y = sqrt(48 - x^2); where x is an interval from 0 to 4sqrt(3)

  26. amistre64
    • 5 years ago
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    4sqrt(3) ------- = 2 sqrt(6); the height for max volume should be 4 sqrt(6) sqrt(2)

  27. amistre64
    • 5 years ago
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    is that the right answer? :)

  28. anonymous
    • 5 years ago
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    h=4 cm, height=2h, therefore, maximum height=8 cm

  29. anonymous
    • 5 years ago
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    thank you! :)

  30. amistre64
    • 5 years ago
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    if I was helpful; youre welcome :)

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