Maximize Vol. of Cylinder w/ Conical End: Solution

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In summary, the conversation discusses the process of maximizing the volume of a projectile in the form of a circular cylinder with one conical end and one flat end, given the surface area. The solution involves using the equations for surface area and volume, along with the relationship between s, r, and theta. The optimal values for r, s, and l depend on a parameter, and the answer is given in terms of a ratio.
  • #1
jesuslovesu
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[SOLVED] Maximizing an Object

Homework Statement


What proportions will maximize the volume of a projectile in the form of a circular cylinder with one conical end and one flat end, if the surface area is given?

(there is a picture given with r being the radius, l being the length of the cylinder and s being the length of the cone(outside edge))

Homework Equations


The Attempt at a Solution



Well I can come up with:
SA = [tex]\pi*r^2+ 2\pi*rl + \pi*rs[/tex]
V = [tex]\pi*r^2l + \pi / 3r^2*s*sin\theta[/tex]

I've tried both using Lagrange multipliers and just using partial derivatives.
but haven't really come up with anything...
(r, l, s)
[tex]<2\pi rl + 2/3\pi*r*s*sin\theta, \pi r^2 + 0, 1/3\pi*r^2*sin\theta> = \lambda*(<2\pi r + 2\pi l + \pi s, 2\pi r, \pi r>[/tex]
Then with using the 3 equations I was getting r = 2l and l = lamda, which isn't the case, so I think I must either need another equation or my equations are wrong?
 
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  • #2
What is wrong here is that sin(theta) is NOT independent of s and r!
Hence, you cannot treat it as a constant.


Use therefore the relationship: [tex]s\sin\theta=\sqrt{s^{2}-r^{2}}[/tex] in your volume formula.
 
  • #3
Thanks for your reply, I am able to get a few relationships between s, r, l, but I can't seem to get the answer yet. (ie r = sqrt(5), s = 3)
I can get
[tex]\sqrt{s^2-r^2} = \frac{2}{3} s [/tex]
[tex]r = \sqrt{5} / 3 s[/tex]
[tex]\lambda = r/2[/tex]
Will my lambda need to change throughout the problem? I seem to consistently be getting the same one, but for whatever reason I can't seem to solve the damn thing.
 
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  • #4
jesuslovesu said:
Thanks for your reply, I am able to get a few relationships between s, r, l, but I can't seem to get the answer yet. (ie r = sqrt(5), s = 3)
I can get
[tex]\sqrt{s^2-r^2} = \frac{2}{3} s [/tex]
[tex]r = \sqrt{5} / 3 s[/tex]
[tex]\lambda = 2/r[/tex]
Will my lambda need to change throughout the problem? I seem to consistently be getting the same one, but for whatever reason I can't seem to solve the damn thing.

I don't get what you are saying.

You have FOUR equations given:
The three "gradient" equations, along with the requirement that the surface area equals some constant C.

Your optimal values of r, s and l (and lambda) will depend upon that C as a parameter.
 
  • #5
My apologies, the book I'm using put the answer in terms of a ratio, I was thinking it was an actual numerical answer.
 
  • #6
jesuslovesu said:
My apologies, the book I'm using put the answer in terms of a ratio, I was thinking it was an actual numerical answer.

Well, a known ratio is an actual numerical answer, it just shows that there are several combinations that maximizes the volume of the object
in question.
It is directly related to the parameter-based approach I'd prefer.
 

FAQ: Maximize Vol. of Cylinder w/ Conical End: Solution

How do you calculate the maximum volume of a cylinder with a conical end?

The maximum volume of a cylinder with a conical end can be calculated using the formula V = πr^2h + (1/3)πr^2h, where V is the volume, r is the radius of the cylinder, and h is the height of the cylinder.

What is the purpose of maximizing the volume of a cylinder with a conical end?

The purpose of maximizing the volume of a cylinder with a conical end is to determine the largest possible volume that can be achieved with a given set of dimensions. This can be useful in various applications, such as designing containers or tanks with maximum capacity.

What are the steps involved in solving for the maximum volume of a cylinder with a conical end?

The steps involved in solving for the maximum volume of a cylinder with a conical end include:

  1. Identifying the given dimensions of the cylinder and cone (radius and height)
  2. Using the formula V = πr^2h + (1/3)πr^2h to calculate the volume
  3. Differentiating the volume formula with respect to the variable (either r or h)
  4. Solving for the critical point by setting the derivative equal to 0 and solving for the variable
  5. Substituting the critical point value into the volume formula to find the maximum volume

What are the applications of maximizing the volume of a cylinder with a conical end in real life?

Maximizing the volume of a cylinder with a conical end has several real-life applications, such as designing containers for storing liquids, optimizing the shape of water tanks, or designing rocket fuel tanks for maximum capacity. It can also be useful in the manufacturing of cones or cylinders for industrial purposes.

Are there any limitations to maximizing the volume of a cylinder with a conical end?

Yes, there are some limitations to maximizing the volume of a cylinder with a conical end. The dimensions of the cylinder and cone must be given or known in order to calculate the maximum volume. Additionally, the formula assumes that the cylinder and cone have a uniform cross-sectional area and do not take into account any irregularities in shape. Moreover, in real-life applications, practical constraints such as cost, material limitations, and structural integrity may also limit the maximum achievable volume.

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