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Max/Min Problems with 3-D Shapes

Date: 03/23/2009 at 13:30:57
From: Jesse
Subject: Minimizing cylinder surface area

I have a math project that I am doing under which I am trying to find 
the dimensions of a cylindrical can that will use the minimal surface
area.

The only time I have maximized area was with problems in which I had
to determine the dimensions that would create the maximum area of
rectangular space.  In that case we used the perimeter as a constraint
equation, solved for one of the variables and then substituted it into
the area equation.  In this case would the volume be my constraint
equation?

I know that volume of a cylinder is pi*r^2h and that the surface area
is 2(pi*r^2)+(2pi*r*h).  Do I solve for r or h?



Date: 03/23/2009 at 15:05:53
From: Doctor Ian
Subject: Re: Minimizing cylinder surface area

Hi Jesse,

You could choose either, and you'd get the same result, since either
choice forces the other.  It would be an interesting exercise for you
to try using both h and r as the variable to substitute for.  What do
you notice? 

Presumably you're given the volume, since the point of a can is to
hold a particular amount of some substance (beans, grease, cola, or
whatever).  Again, the idea is the same as for the 2-D problems you've
solved in the past. 

Note that using numbers right away tends to make things more 
confusing.  It's simpler if you just use variables everywhere.

That is, if you have a cylinder, the volume and surface area are 
given by

  V = pi r^2 h

and

  A = 2 pi r^2 + 2 pi r h

    = 2 pi (r^2 + r h)

Suppose you're given a volume, and you want to minimize surface area.
You can solve for height in terms of volume and radius, 

             V = pi r^2 h

  V / (pi r^2) = h

Now you can substitute into the equation for area:

  A = 2 pi (r^2 + r [V / (pi r^2)])

Now, note that you have two variables here: V and r.  As soon as you
specify a volume, you have only one variable, so you can apply the
optimization techniques you've been learning.  Nice, huh? 

On the other hand, suppose you're given a particular amount of surface
area to use, and you want to maximize the volume.  You might start by
solving for height in terms of area and radius, 

                       A = 2 pi (r^2 + r h)
                         .
                         .
  [A / (2 pi) - r^2] / r = h
  
Then you could substitute into the equation for volume:

   V = pi r^2 ([A / (2 pi) - r^2] / r)

and you're in the same boat.  You have two variables (A and r), but as
soon as you specify an area, you have only one variable. 

The numbers just make it harder to see what's going on.  If you leave
everything unspecified until the end, you can just solve each kind of
problem once, for all time. 

Does this make sense? 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 03/23/2009 at 17:15:43
From: Jesse
Subject: Thank you (Minimizing cylinder surface area)

Thank you very much, this solved my issue completely.
Associated Topics:
High School Calculus

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