INSCRIBING A CYLINDER INTO A CONE

[xokp1026xo]

HEY... I need help with this question>>>

find the dimension of the right circular cylinder of largest possible volume which can be inscribed in a right circular cone having radius 3 in. and height of 6tn.


ok I know how to start the problem.. well the formula for volume of a cylinder is V=pie*radius square*hight

I also know that the formula for a cone is V=Pie*radius squared*hight divided by 3
well do the substitution I found that the volume of the cone is 18pie.

next I know I have to do something to the cylinder formula in order to get rid of one of the two variable... and then find the derivate.. all I need help with is how to solve for in terms of one of the variable in the cylinder formula..


please .. any help will be greatly appreciated...

 

[Deleted member 4993]

HEY... I need help with this question>>>

find the dimension of the right circular cylinder of largest possible volume which can be inscribed in a right circular cone having radius 3 in. and height of 6tn.


ok I know how to start the problem.. well the formula for volume of a cylinder is V=pie*radius square*hight

I also know that the formula for a cone is V=Pie*radius squared*hight divided by 3
well do the substitution I found that the volume of the cone is 18pie.

next I know I have to do something to the cylinder formula in order to get rid of one of the two variable... and then find the derivate.. all I need help with is how to solve for in terms of one of the variable in the cylinder formula..


please .. any help will be greatly appreciated...

You need to draw a sketch

Draw a circle to represent the sphere.[--ouch -- you need a cone -- somehow I thought you are dealing with a sphere. However, the principle remains the same. Now you need to draw an isoscales triangle to represent the cone]

Draw a rectangle inside - to represent the cylinder - with the vertices on the perimeter. The width (of the rectangle) will represent the diameter of the cylinder and the length will be the height of the cylinder

Now use Pythagorus to relate the radius of the sphere, radius of the cylinder and (half) height of the cylinder.[for a cone you'll have similar triangles and use the laws of proportionality instead]

Then continue...

 

[Dr. Flim-Flam]

Volume of a cylinder = ?r²h. Let r= radius of cylinder and h = height of cylinder.

In cross-section, we have due to similair triangles: 6/3 = (6-h)/r, h = 6-2r.

V of cylimder = ?r²h = ?* r²*(6-2r) ; dV/dr = ?[12r - 6r²]. Setting the slope = to zero, we get

?[6r(2-r)] = 0, r=0,r=2. Ergo h =6-2r, h = 2 .

Therefore the max dimensions of cylinder are r = 2 , h = 2 and volume = 8?.

Note: Let R and H equal the radius and height of the cone respectively and r and h equal

the radius and height of the cylinder respectively; then the max right circular cylinder that can be inscribed

into the right circular cone is r = 2R/3 and h = H/3. In the above case, r = 2 and h = 2.

Post Script: What is with the "pie"?