MHB Maximizing the weight of a cylinder cut from a sphere

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The discussion focuses on determining the weight of the largest cylinder that can be cut from a sphere weighing P kg. It establishes that the relationship between the mass, volume, and density is crucial, leading to the formula for the cylinder's weight as a function of the sphere's weight. Participants suggest using geometric relationships and calculus to derive the volume of the cylinder in terms of its height and radius, ultimately aiming to maximize this volume. The critical point for maximizing the cylinder's volume is found, and the final weight of the cylinder is expressed in relation to the sphere's weight. The conversation concludes with a participant successfully arriving at the solution.
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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
 
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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?
 
Re: max and min 4

MarkFL said:
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
 
Re: max and min 4

leprofece said:
I can do that by pitagorean theoreme

Yes, that's correct. What is the relationship between mass, volume and mass density?
 
Re: max and min 4

P or mass = d.v
 
Re: max and min 4

leprofece said:
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?
 
Re: max and min 4

MarkFL said:
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 don't know how to eliminate r and h to get (sqrt of 3)m/3 that is my book answer
 
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.
 
Re: max and min 4

ohhh I can't find this relationship really I give up sorry
I would appreciate your helping
 
  • #10
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
 

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  • #11
Re: max and min 4

MarkFL said:
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:
  • #12
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.

$$w_C$$ = weight of the cylinder

$$w_S=P$$ = weight of the sphere

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

$$\rho=\frac{P}{V_S}=\frac{w_C}{V_C}$$

Hence:

$$w_C=\frac{V_C}{V_S}P$$

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:

$$V_C=\pi r^2h$$

Subject to the constraint:

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

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 $r^2$:

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

Now, differentiate this with respect to $h$ and equate the result to zero to determine the critical value(s).
 
  • #13
Re: max and min 4

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

$$w_C$$ = weight of the cylinder

$$w_S=P$$ = weight of the sphere

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

$$\rho=\frac{P}{V_S}=\frac{w_C}{V_C}$$

Hence:

$$w_C=\frac{V_C}{V_S}P$$

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:

$$V_C=\pi r^2h$$

Subject to the constraint:

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

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 $r^2$:

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

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:
 
  • #14
Re: max and min 4

leprofece said:
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:

$$w_C=\frac{V_C}{V_S}P$$
 
  • #15
Re: max and min 4

MarkFL said:
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:

$$w_C=\frac{V_C}{V_S}P$$

and how does Vs remains??
must I substitute this value too there??
 
  • #16
Re: max and min 4

leprofece said:
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.
 
  • #17
Re: max and min 4

MarkFL said:
$V_S$ is the volume of the sphere, so you want to substitute its formula there.

OK I GOT THE ANSWER THANKS A LOT(Angel)
 

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