# Circular Cylinder Problem

#### leprofece

##### Member
In a cone circular line of 15 cm in height and radius 5 cm fits a body cylindrical topped by 1 hemisphere tangent to the base of the cone. Calculate the height and radius of Ia part cylindrical if the volume of the registered body is the largest possible

Answer R = H = 3 cm

V= pir^2h/3 cone
V Hemisfere = 2pir^3/3

I dont know how to solve h if in the other there is no H

#### Sudharaka

##### Well-known member
MHB Math Helper
Re: max and min 288

In a cone circular line of 15 cm in height and radius 5 cm fits a body cylindrical topped by 1 hemisphere tangent to the base of the cone. Calculate the height and radius of Ia part cylindrical if the volume of the registered body is the largest possible

Answer R = H = 3 cm

V= pir^2h/3 cone
V Hemisfere = 2pir^3/3

I dont know how to solve h if in the other there is no H
Hi leprofece,

I am not sure I understand your problem. Did you translate this from another language?

#### MarkFL

Staff member
Re: max and min 288

Hi leprofece,

I am not sure I understand your problem. Did you translate this from another language?
I believe the problem to be:

A right circular cylinder capped with a hemisphere at one end (we'll call this object "the body") is enclosed by a right circular cone of radius 5 cm and height 15 cm, such that the base of the cone and the hemispherical cap are tangent to one another and the cone and "the body" share the same axis of symmetry. Here is a cross-section of the two bodies, through this common axis of symmetry:

Find the dimensions of "the body" which maximizes its volume.

leprofece, does this sound correct to you?

#### leprofece

##### Member
Yes that is right
it is correct yoour solving
may you help me??

#### MarkFL

Staff member
Yes that is right
it is correct yoour solving
may you help me??
First, can you state the objective function? Then can you state the constraint using the cross section? I would approach the constraint using coordinate geometry.

#### leprofece

##### Member
It is complicate
From the figure volume of the figure introduced
h (r^2) = V
It remains a triangle maybe it applies Thales
to get the another equation 2r/h = 2h/x

#### MarkFL

Staff member
It is complicate
From the figure volume of the figure introduced
h (r^2) = V
It remains a triangle maybe it applies Thales
to get the another equation 2r/h = 2h/x
The volume of the body is the sum of that of a cylinder and a hemisphere.

#### leprofece

##### Member
The volume of the body is the sum of that of a cylinder and a hemisphere.
ok it must be one of the two equations
pir2*h+ 2/3pir3

and the another must be by thales
????'

#### MarkFL

Staff member
I would use coordinate geometry for the constraint. The cross-section itself has bilateral symmetry and so you really only need consider one side of the cone. I would put the vertex of the cone at the origin, and then find the line that coincides with the slant edge. Then label everything you know and see if you can get the height of the cylindrical portion of the body in terms of the radius of the body, and other constants.

#### leprofece

##### Member
if so Point A is (0, 0) y b) (r,h)
m = h/r

the line wouuld be y-h = h/r(x-r)
y = hx -hr +h
y = hx-h(r-1)/r

maybe i must solve in the volume equation for h
V= pir^2h/3 + 2 pir^3/3

It is easier solve from here and now who are y and x ?

#### MarkFL

Staff member
Consider this diagram:

Now, can you find $h$ in terms of the constants $H$ and $R$ and the variable $r$?

#### leprofece

##### Member
Consider this diagram:

View attachment 2082

Now, can you find $h$ in terms of the constants $H$ and $R$ and the variable $r$?
y = hx/r is ready so
the another part of the triangle is thales
h-r/R = H/r
So I must solve from here must not I ???

#### MarkFL

Staff member
y = hx/r is ready so
the another part of the triangle is thales
h-r/R = H/r
So I must solve from here must not I ???
That's not what I get. I used the fact that:

$$\displaystyle a+h+r=H$$

and from similarity or the equation of the line, we find:

$$\displaystyle a=\frac{H}{R}r$$

Now, substitute for $a$ into the first equation, and solve for $h$. What do you find?

#### leprofece

##### Member
That's not what I get. I used the fact that:

$$\displaystyle a+h+r=H$$

and from similarity or the equation of the line, we find:

$$\displaystyle a=\frac{H}{R}r$$

Now, substitute for $a$ into the first equation, and solve for $h$. What do you find?
I Find
Hr/R+r + h = H
Putting the data problem
5H/R+5+15 = H
H-5H/R = 20
HR -5H = 20 R
H(R-5) = 20 R
H=20R/R-5
derivating
I dont get H=R = 3 that is the book answer ·

#### MarkFL

Staff member
Okay, we have:

$$\displaystyle \frac{H}{R}r+h+r=H$$

Solving for $h$, we find:

$$\displaystyle h=H-r\left(1+\frac{H}{R} \right)$$

Now, our objective function is:

$$\displaystyle V(r,h)=\pi r^2h+\frac{2}{3}\pi r^3$$

Substituting for $h$, we obtain:

$$\displaystyle V(r)=\pi r^2\left(H-r\left(1+\frac{H}{R} \right) \right)+\frac{2}{3}\pi r^3$$

Distributing and combining like terms, we get:

$$\displaystyle V(r)=\pi Hr^2+\pi r^3\left(\frac{2}{3}-\left(1+\frac{H}{R} \right) \right)$$

This is the function you want to maximize.