Optimization cylindrical can Problem

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Homework Help Overview

The discussion revolves around an optimization problem involving a cylindrical can with a fixed volume. The original poster attempts to prove that the height must equal the diameter to minimize material usage.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the height and diameter of the cylinder, with some suggesting they are equal while others question this assumption. There are attempts to derive expressions for volume and surface area, and to understand how calculus might be applied to find a minimum.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the need for separate expressions for volume and surface area, and the necessity of deriving these to proceed with the optimization.

Contextual Notes

There is a focus on the assumption that height and diameter are equal, which some participants challenge. The problem requires showing relationships mathematically rather than assuming them.

Tido611
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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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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
 
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.
 
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.
 
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
 
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?
 
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.
 
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?
 

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