Question

WHAT IS THE MAXIMUM VOLUME OF CYLINDER?

Answer 1

Depends on the raidus and height.

Answer 2

A circular cone has radius of 4 cm , height of 12 cm ( the cylinder is upstanding inside of the circular cone)

Answer 3

that was fast

Answer 4

What is the volume of the cylinder?

Answer 5

OH I understand your problem now!

Answer 6

that is the question??of the question???

Answer 7

So basically you have a cone, and there is a cylinder inside, and you want to find max voulme possible of the cylinder?

Answer 8

yup

Answer 9

it should be a formula to solve it

Answer 10

Hmm, let me think about it then.

Answer 11

Let's say the height of the cylinder is x and it's radius is y. From the top of the cylinder up to the top of the cone, there will be a smaller cone.
If we draw a section through the center of the cone, perpendicular on the base, the image will look like a rectangle (the cylinder), inside a triangle (the cone). If the bottom-left angle of the figure is alpha, we can say that tan(alpha)=h/r.
Above the rectangle there will be a smaller triangle. Its bottom-left angle is again alpha, but its tangent is now tan(alpha)=(h-x)/y.
h/r=(h-x)/y means y=(h-x)r/h
The volume of the cylinder is:
V(x) = πy²x = π(h-x)²r² / (h²x) = (πr²/h²) · ((h-x)²/x)
Its minimums and maximums are given by deriving it
V'(x) = (πr²/h²) · (1/x²) · ( (x ·2(h-x) ) - (h-x)² )
V'(x) = ( πr²/(h²x²) ) · (h-x) (2x-h+x)
V'(x) = (h-x)(3x-h) · πr²/(h²x²)
The condition for minimums/maximums is V'(x)=0, or given that πr²/(h²x²) is always non-null, positive, (h-x)(3x-h)=0.
This has two solutions x=h and x=h/3, but only the last one makes sense. (the cone should be taller than the cylinder, i.e. x<h).
If x=h/3, y = (h-x)r/h = (2h/3)r/h = 2r/3, and the maximum volume is V = πy²x = π(2r/3)²(h/3)= (4π/27)·r²h.

Answer 12

thanks for the basics it will really help me ...
thanks alot