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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?
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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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1whats the problem? we dont all see what you see :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Find 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?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1whose radius? cyl or sphere?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1we can take the cross section and just figure out the rectangle in a circle... right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, 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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1do we want to maximize the area of the cyl? or just the height?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Like this: sorry for the cheesy sketch =P

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1area...volume; same diff :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1its 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In 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?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1this is usually associated with the question: What is the largest volume right sylindar that can fit inside a sphere of radius (#)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the shapes I could deal with; its the missing information to make the question plausible thats getting me :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea, the solution in front of me seems pretty simple, but when i try to solve myself I always end up messing up somewhere...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1What 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there is no missing info, from the sketch you get a constraint that's like this (r^2=48h^2) and you use that for r^2 in your function which is the volume of the cylinder (V= pie*r^2h)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1So it IS volume we want to maximize....right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then 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.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1which of these is the radius of the sphere? \[\sqrt[4]{3} \leftarrow \rightarrow 4\sqrt{3}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1good :) then we can work this easier :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the end reply for h=4 cm, since h was divided into two, the height would be 2h, therefore 8 cm =)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1it pretty much boils down to this; what is the largest area rectangle you can fit under a quarter top circle..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the 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)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.14sqrt(3)  = 2 sqrt(6); the height for max volume should be 4 sqrt(6) sqrt(2)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1is that the right answer? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0h=4 cm, height=2h, therefore, maximum height=8 cm

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1if I was helpful; youre welcome :)
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