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Maximizing the weight of a cylinder cut from a sphere

leprofece

Member
Jan 23, 2014
241
A sphere weighs P kg what is the weight of the higher straight circular cylinder that can cut from the sphere?

Answer sqrt(3)P/3
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: max and min 4

Assuming the mass density of the sphere is constant throughout, how are mass and volume related?

I suggest drawing a diagram of a cross section of the two objects through the axis of symmetry of the cylinder and try to find a relationship between the radius and height of the cylinder with the radius of the sphere. What do you find?
 

leprofece

Member
Jan 23, 2014
241
Re: max and min 4

Assuming the mass density of the sphere is constant throughout, how are mass and volume related?

I suggest drawing a diagram of a cross section of the two objects through the axis of symmetry of the cylinder and try to find a relationship between the radius and height of the cylinder with the radius of the sphere. What do you find?
I can do that by pitagorean theoreme
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: max and min 4

I can do that by pitagorean theoreme
Yes, that's correct. What is the relationship between mass, volume and mass density?
 

leprofece

Member
Jan 23, 2014
241
Re: max and min 4

P or mass = d.v
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: max and min 4

P or mass = d.v
Yes, and if the density is constant, then we know mass and density are proportional to one another, and so we need only compute the ratio of the volume of the cylinder to that of the sphere, and multiply this by the mass of the sphere to get the mass of the cylinder.

So, what is the relationship between the dimensions (radius and height) of the cylinder with the radius of the sphere?

What is your objective function?
 

leprofece

Member
Jan 23, 2014
241
Re: max and min 4

Yes, and if the density is constant, then we know mass and density are proportional to one another, and so we need only compute the ratio of the volume of the cylinder to that of the sphere, and multiply this by the mass of the sphere to get the mass of the cylinder.

So, what is the relationship between the dimensions (radius and height) of the cylinder with the radius of the sphere?

What is your objective function?
m= d.v
and sphre volume is 4pir3/3
Then m = d(4pir3/3)
r2= R2+ H2
i tried to derive v respect to r
m= d (4pi(R2+ H2)3
looking at the answer i dont know how to eliminate r and h to get (sqrt of 3)m/3 that is my book answer
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: max and min 4

Look at a cross-section of the sphere and cylinder through their mutual center. You should find that your relationship (using the Pythagorean theorem) between the radius $R$ of the sphere and the radius $r$ and $h$ of the cylinder is incorrect. You have one side of the right triangle wrong.
 

leprofece

Member
Jan 23, 2014
241
Re: max and min 4

ohhh I cant find this relationship really I give up sorry
I would appreciate your helping
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: max and min 4

Here is a diagram...do you see from where I obtained the measures of the 3 sides of the right triangle?

leprofece3.jpg
 

leprofece

Member
Jan 23, 2014
241
Re: max and min 4

Here is a diagram...do you see from where I obtained the measures of the 3 sides of the right triangle?

View attachment 1960
lets see
m= d.v
and sphre volume is 4pir3/3
Then m = d(4pir3/3)
r2= R2- (H/2)2
i tried to derive v respect to r
m= d (4pi(R2- (H/2)2])3/3
now can i derive from here??
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: max and min 4

I would just maximize the volume of the cylinder first, and then use the relationship between weight density and weight.

\(\displaystyle w_C\) = weight of the cylinder

\(\displaystyle w_S=P\) = weight of the sphere

Thus, we may state, given that both objects share the same weight density:

\(\displaystyle \rho=\frac{P}{V_S}=\frac{w_C}{V_C}\)

Hence:

\(\displaystyle w_C=\frac{V_C}{V_S}P\)

You already know the volume of the sphere, so all you need now is the volume of the cylinder. So, our objective function is the volume of the cylinder:

\(\displaystyle V_C=\pi r^2h\)

Subject to the constraint:

\(\displaystyle r^2+\left(\frac{h}{2} \right)^2=R^2\)

So, using the constraint, we may write the volume of the cylinder in terms of 1 variable, and it will be simpler to substitute for $r^2$:

\(\displaystyle V_C(h)=\pi\left(R^2-\left(\frac{h}{2} \right)^2 \right)h=\pi R^2h-\frac{\pi}{4}h^3\)

Now, differentiate this with respect to $h$ and equate the result to zero to determine the critical value(s).
 

leprofece

Member
Jan 23, 2014
241
Re: max and min 4

I would just maximize the volume of the cylinder first, and then use the relationship between weight density and weight.

\(\displaystyle w_C\) = weight of the cylinder

\(\displaystyle w_S=P\) = weight of the sphere

Thus, we may state, given that both objects share the same weight density:

\(\displaystyle \rho=\frac{P}{V_S}=\frac{w_C}{V_C}\)

Hence:

\(\displaystyle w_C=\frac{V_C}{V_S}P\)

You already know the volume of the sphere, so all you need now is the volume of the cylinder. So, our objective function is the volume of the cylinder:

\(\displaystyle V_C=\pi r^2h\)

Subject to the constraint:

\(\displaystyle r^2+\left(\frac{h}{2} \right)^2=R^2\)

So, using the constraint, we may write the volume of the cylinder in terms of 1 variable, and it will be simpler to substitute for $r^2$:

\(\displaystyle V_C(h)=\pi\left(R^2-\left(\frac{h}{2} \right)^2 \right)h=\pi R^2h-\frac{\pi}{4}h^3\)

Now, differentiate this with respect to $h$ and equate the result to zero to determine the critical value(s).
ok I got +/- 2Rsqrt3/3 as critical point
Now we need to put that respect to P:confused:
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: max and min 4

ok I got +/- 2Rsqrt3/3 as critical point
Now we need to put that respect to P:confused:
First, I would demonstrate that this critical value is at a relative maximum. My preference here would be the second derivative test.

Then, you want to evaluate $V_C$ at this critical value, and then substitute that into the formula:

\(\displaystyle w_C=\frac{V_C}{V_S}P\)
 

leprofece

Member
Jan 23, 2014
241
Re: max and min 4

First, I would demonstrate that this critical value is at a relative maximum. My preference here would be the second derivative test.

Then, you want to evaluate $V_C$ at this critical value, and then substitute that into the formula:

\(\displaystyle w_C=\frac{V_C}{V_S}P\)
and how does Vs remains??
must I substitute this value too there??
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: max and min 4

and how does Vs remains??
must I substitute this value too there??
$V_S$ is the volume of the sphere, so you want to substitute its formula there.
 

leprofece

Member
Jan 23, 2014
241
Re: max and min 4

$V_S$ is the volume of the sphere, so you want to substitute its formula there.
OK I GOT THE ANSWER THANKS A LOT(Angel)