Optimization cylindrical can Problem

In summary: Yes, you would set it equal to 0. In summary, you cannot make a cylindrical can of volume K cubic units using a minimum of material if the height is greater than the diameter.
  • #1
Tido611
79
0
Prove that any cylindrical can of volume K cubic units that is to be made using a minimum of material must have the height equal to the diameter.
 
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  • #2
Hi,

I've moved your thread to the homework help section. In order to receive help you must show how you started this problem.

Thanks,

Tom
 
  • #3
all I've got so far is 2(pi(1/2x)^2)+(x^2pi) and all that explains is the area of any can given the height/diameter. But i don't know how to prove it with calculus.
 
  • #4
Your equation doesn't make any sense to me. For starters it only has one variable, but it should have two.

Let [itex]d[/itex] be the diameter and let [itex]h[/itex] be the height. Can you state expressions for the total volume [itex]V[/itex] and the total surface area [itex]S[/itex] in terms of [itex]d[/itex] and [itex]h[/itex]?

If so then you're almost home.
 
  • #5
there is no reason to separate d and h from each other because they are the same value d=h=x

and the equation for volume is (pi)r^2x
 
  • #6
there is no reason to separate d and h from each other because they are the same value d=h=x
No they are not:

(1) Firstly, the diameter of a cylinder is an entirely different thing than the height of a cylinder.

(2) Secondly, nothing in the optimization problem even (directly) suggests that d and h should be equal... it says that the answer should have d and h equal.


If you're still confused by (2), consider this analogy:

You are asked to prove or disprove that the only solution to the equation [itex]x^2 + 3x + 2 = 0[/itex] is [itex]x = 2[/itex]. How would you do it?
 
  • #7
Tido611 said:
there is no reason to separate d and h from each other because they are the same value d=h=x

*thumps head*

You're supposed to show that, not assume it!

and the equation for volume is (pi)r^2x

OK, how about surface area? You're going to need that, too.
 
  • #8
ok so the whole point of this problem is to solve everything and then compare d to h in the end

so the surface area being 2pi*r*h + 2pi*r^2, would you find the derivative of this and set it equal to 0?
 

What is the "Optimization cylindrical can Problem"?

The Optimization cylindrical can Problem is a mathematical problem that involves finding the dimensions of a cylindrical can that will maximize its volume given a fixed amount of material. It is a common problem in optimization and engineering.

What factors are involved in the Optimization cylindrical can Problem?

The main factors involved in the Optimization cylindrical can Problem are the height and radius of the can, and the amount of material available to make the can. Other factors may include cost, strength, and practicality.

What is the formula for calculating the volume of a cylindrical can?

The formula for calculating the volume of a cylindrical can is V = πr2h, where V is volume, r is the radius, and h is the height.

What are the steps for solving the Optimization cylindrical can Problem?

The steps for solving the Optimization cylindrical can Problem are: 1) Define the objective function (in this case, the volume of the can); 2) Identify any constraints (e.g. the amount of material available); 3) Use calculus to find the critical points of the objective function; 4) Evaluate the critical points to determine the maximum volume; 5) Check the endpoints of the possible range of values for the dimensions to ensure the maximum volume has been found.

What are some real-world applications of the Optimization cylindrical can Problem?

The Optimization cylindrical can Problem has applications in various industries, such as packaging, manufacturing, and engineering. For example, it can be used to determine the dimensions of a soda can to minimize material waste and production costs. It can also be applied to design more efficient fuel tanks or storage containers.

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