Standard kilogram physics help

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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.
 

HallsofIvy

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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.
 
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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?
 
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Second solution: when A = 2[pi] r^2 + 2[pi] rh
 

HallsofIvy

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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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