## anonymous 5 years ago Does anyone know any tips to follow for optimization problems and how to deal with the different shapes in them?

1. amistre64

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

2. 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?

3. amistre64

4. anonymous

sphere

5. amistre64

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

6. 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

7. amistre64

like this right?

8. amistre64

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

9. anonymous

Like this: sorry for the cheesy sketch =P

10. amistre64

area...volume; same diff :)

11. 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

12. 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?

13. amistre64

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

14. amistre64

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

15. 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...

16. 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?

17. 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)

18. amistre64

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

19. 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.

20. amistre64

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

21. anonymous

second one

22. amistre64

good :) then we can work this easier :)

23. anonymous

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

24. amistre64

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

25. 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)

26. amistre64

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

27. amistre64

is that the right answer? :)

28. anonymous

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

29. anonymous

thank you! :)

30. amistre64

if I was helpful; youre welcome :)