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

- anonymous

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- amistre64

whats the problem? we dont all see what you see :)

- anonymous

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?

- amistre64

whose radius? cyl or sphere?

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## More answers

- anonymous

sphere

- amistre64

we can take the cross section and just figure out the rectangle in a circle... right?

- anonymous

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

- amistre64

like this right?

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- amistre64

do we want to maximize the area of the cyl? or just the height?

- anonymous

Like this: sorry for the cheesy sketch =P

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- amistre64

area...volume; same diff :)

- amistre64

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

- anonymous

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?

- amistre64

this is usually associated with the question: What is the largest volume right sylindar that can fit inside a sphere of radius (#)

- amistre64

the shapes I could deal with; its the missing information to make the question plausible thats getting me :)

- anonymous

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

- amistre64

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?

- anonymous

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)

- amistre64

So it IS volume we want to maximize....right?

- anonymous

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.

- amistre64

which of these is the radius of the sphere?
\[\sqrt[4]{3} \leftarrow \rightarrow 4\sqrt{3}\]

- anonymous

second one

- amistre64

good :) then we can work this easier :)

- anonymous

the end reply for h=4 cm, since h was divided into two, the height would be 2h, therefore 8 cm =)

- amistre64

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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- amistre64

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)

- amistre64

4sqrt(3)
------- = 2 sqrt(6); the height for max volume should be 4 sqrt(6)
sqrt(2)

- amistre64

is that the right answer? :)

- anonymous

h=4 cm, height=2h, therefore, maximum height=8 cm

- anonymous

thank you! :)

- amistre64

if I was helpful; youre welcome :)

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