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Oil Can Dimensions


Date: 12/11/2001 at 23:34:29
From: Shannon 
Subject: Difficult word problem

An oil can is in the form of a cylinder with a base 5cm in radius. If 
the oil can holds one liter, how tall is it?

I have already figured out this part of the question; it is the next 
part that is giving me all the trouble.

What are the dimensions of an oil can with a one-liter capacity that 
uses the least amount of tin? 

The prof. swears that no calculus is needed to complete the problem, 
but I beg to differ. Thank you.
Shannon


Date: 12/12/2001 at 00:31:38
From: Doctor Jeremiah
Subject: Re: Difficult word problem

Hi Shannon,

Here is the method for getting minimums and maximums:

The most important thing in order to find minumum area is to get an 
equation for the area that has only one independant variable.

The minimum surface area will be a can with an unknown radius and 
height. So the values that you have will not work for the minimum and 
we will have to find new ones. (The ones you have are for a can that 
does not have the minimum area.)

First you find an equation for the surface area. Since it's a right
circular cylinder the surface area is:

   A = 2 * area of base + area of sides
   A = 2 * area of circle + circumferecece of circle * height
   A = 2 * Pi*radius^2 + 2*Pi*radius*height

But that is not all you need. There are too many variables. We need to 
get rid of the height. The other clue is that the can always holds one 
litre, which means we need an equation for volume:

           V = area of base * height
   1000 cm^3 = 1 L = area of circle * height
        1000 = Pi*radius^2 * height

Now we can use the equation for volume to determine the height and 
substitute that into the equation for surface area.

        1000 = Pi*radius^2 * height
      height = 1000 / Pi*radius^2

Okay, we have a value for the height in terms of radius. And if we 
substitute that into the area:

   A = 2 * Pi*radius^2 + 2*Pi*radius*(1000 / Pi*radius^2)
   A = 2 * Pi*radius^2 + 2000/radius

Now we have an equation for area that will work, because it has only 
one independant variable (the radius).

Think about what the graph of the area will look like. The minimum 
will be at the bottom of a dip where the slope changes from negative 
to positive. The absolute bottom of the dip will have a slope of zero.

So what we need to do is find out where the slope of this equation is 
equal to zero. You could actually draw the graph and figure it out.  
Thats why the professor said you didn't need calculus. Just graph this 
equation and look for the minimum:

   A = 2 * Pi*radius^2 + 2000/radius

But since I am not going to draw a graph, I *will* use calculus to 
calculate it this time. Taking the derivative gives the slope, so all 
we have to do is differentiate the equation for area to get the slope 
of the area:

              A = 2 * Pi*radius^2 + 2000/radius

If we differentiate it:

 d/dradius( A ) = d/dradius( 2 * Pi*radius^2 + 2000/radius )

After you take the derivative you end up with:

     dA/dradius = 4*Pi*radius - 2000/radius^2

Now remember we said that the minimum was when the slope was 0?

      slope = 0 = 4*Pi*radius - 2000/radius^2

And if you solve that for minimum radius:

     5.41926 cm = radius

Just to show you it is the right answer, here are some calculations of 
points near the minimum:

     radius = 5.40 cm     A = 553.588 cm^2
     radius = 5.41 cm     A = 553.583 cm^2
     radius = 5.41926 cm  A = 553.581 cm^2  <==  minimum
     radius = 5.43 cm     A = 553.583 cm^2
     radius = 5.44 cm     A = 553.589 cm^2

Let me know if you would like more details on the calculus part.

- Doctor Jeremiah, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus
High School Geometry
High School Higher-Dimensional Geometry
High School Practical Geometry

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