Standard kilogram physics help

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Discussion Overview

The discussion revolves around the mathematical problem of determining the dimensions of a circular cylinder that minimize the surface area while maintaining a fixed volume. The context includes references to a textbook problem and involves exploring the relationship between the height and radius of the cylinder.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant requests hints on how to show that a cylinder's height equal to its radius minimizes surface area for a fixed volume.
  • Another participant outlines the standard approach, suggesting the use of derivatives to find the minimum surface area based on the formulas for surface area and volume.
  • A participant presents two different results regarding the conditions for minimum surface area: one stating that minimum area occurs when height equals radius, and another suggesting it occurs when height equals twice the radius.
  • Clarifications are made about the surface area formulas, with one participant emphasizing the correct interpretation of the surface area of a solid cylinder versus an open-ended can.

Areas of Agreement / Disagreement

Participants express disagreement regarding the conditions for minimum surface area, with one proposing that height should equal radius and another suggesting height should equal twice the radius. The discussion remains unresolved as to which condition is correct.

Contextual Notes

Participants have not reached consensus on the correct relationship between height and radius for minimizing surface area, and there are differing interpretations of the surface area formulas based on the context of the problem.

Hyperreality
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The standard kilogram is in the shape of a circular cylinder with its height equal to its dimager. Show that, for a circular cylinder of fixed volume, this equality gives the smallest surface area, thus minimizing the effects of surface contamination and wear.

Now some of you might have seen this problem in Fundamentals of Physics Extended Fifth Edition. Please give me some hints.
 
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The problem is to show that making the height of a cylinder equal to its radius minimizes the surface area (for a given volume). The standard way of doing such a problem is to write the formula for the quantity, then take the derivative.

A cylinder of radius r and height h has three surfaces: the top and bottom, each of area [pi]r2, and the lateral surface which has area 2[pi]rh ("uncurl" the lateral surface and it is a rectangle with area the length of the circumferences time the height). The total area is A= 2[pi]r2+ 2[pi]rh= 2[pi](r2+ rh). The Volume is the area of the base time the height:
V= [pi]r2h so h= V/([pi]r2) and we can write
the surface area as A= 2[pi](r2+ V/([pi]r)) (V is a constant).

Differentiate with respect to r, set equal to 0 and solve for r.
You will get a formula for r that depends on V. Substitute for V with the formula above and see what happens.
 
Yes, I obtained to results from two different surface areas.

First solution:
A minimum area is obtained when h = r for A = [pi] r^2 + 2[pi] rh.

Second solution:
A minimum surface area is obtained when h = 2r.

Can anyone please explain the result to me?
 
Second solution: when A = 2[pi] r^2 + 2[pi] rh
 
A= 2[pi]r2+ 2[pi]rh is the surface area of a cylinder:
the "lateral surface" area is 2[pi]rh and each end has area [pi]r2.

A= [pi]r2+ 2[pi]rh would be the area of the lateral surface and ONE end. If you were doing a problem concerning a "can" open at one end you would use that but "the standard kilogram in the shape of a circular cylinder" is a solid cylinder so there is no reason to use that.
 

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