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Mathematics 14 Online
OpenStudy (anonymous):

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

OpenStudy (amistre64):

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

OpenStudy (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?

OpenStudy (amistre64):

whose radius? cyl or sphere?

OpenStudy (anonymous):

sphere

OpenStudy (amistre64):

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

OpenStudy (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

OpenStudy (amistre64):

like this right?

OpenStudy (amistre64):

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

OpenStudy (anonymous):

Like this: sorry for the cheesy sketch =P

OpenStudy (amistre64):

area...volume; same diff :)

OpenStudy (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

OpenStudy (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?

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (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...

OpenStudy (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?

OpenStudy (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)

OpenStudy (amistre64):

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

OpenStudy (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.

OpenStudy (amistre64):

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

OpenStudy (anonymous):

second one

OpenStudy (amistre64):

good :) then we can work this easier :)

OpenStudy (anonymous):

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

OpenStudy (amistre64):

it pretty much boils down to this; what is the largest area rectangle you can fit under a quarter top circle..

OpenStudy (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)

OpenStudy (amistre64):

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

OpenStudy (amistre64):

is that the right answer? :)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

thank you! :)

OpenStudy (amistre64):

if I was helpful; youre welcome :)

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