Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Maximum Surface Area for Total Edge Length

Date: 07/14/2002 at 04:31:16
From: Peter Koloszar
Subject: Calculus-max. and min.

Dear Sir, 

Please help me solve the following problem: 

  A piece of wire of total length L units is used to form 
  the nine edges of a prism whose ends are equilateral 
  triangles and whose other faces are rectangles. Show that 
  the total surface area of the prism cannot exceed  
  L^2/[2(12 - sqrt(3))] square units.

Thank you for your help. 
Peter Koloszar.


Date: 07/16/2002 at 02:57:46
From: Doctor Jeremiah
Subject: Re: Calculus-max. and min.

Hi Peter,

The edges of the prism add up add to L because that's how 
long the wire is.

If the prism has edges at the ends of length "a" units 
(because they are equilateral) and edges along the prism of 
length "b" units then the prism would look like:

           +
         /   \
       +-------+
       |   |   |
       |   |   |
       |   |   |
       |   |   b
       |   |   |
       |   +   |
       | /   \ |
       +---a---+

The total length of the edges is:

   L = 6a + 3b

And the surface area is:

   S = 3(rectangle_area) + 2(triangle_area)

     = 3ab + a sqrt(3a^2/4)

So now we have two equations with a and b as unknowns.
We can get rid of one unknown by substitution:

   L = 6a + 3b
   b = L/3 - 2a

   S = 3ab + a sqrt(3a^2/4)   <===    b = L/3 - 2a

And then we can simplify the equation for surface area:

   S = 3ab + a sqrt(3a^2/4)   <===    b = L/3 - 2a

   S = 3a(L/3 - 2a) + a sqrt(3a^2/4)

   S = aL - 6a^2 + a^2 sqrt(3/4)

   S = aL + a^2[sqrt(3/4) - 6]

   S = aL + 2 a^2[sqrt(3/4) - 6]/2

   S = aL + a^2[2 sqrt(3/4) - 12]/2

   S = aL + a^2[sqrt(3) - 12]/2

Now we have an equation for surface area.  But we don't
want that.  We want the MAXIMUM surface area.  To find the
maximum we could graph the equation and look at the graph.
The maximum is where the slope of the graph changes from
positive to negative.  To do that it must go through zero.
When the slope is zero that is the maximum surface area.
So we need to find the slope.  We can do that with a
derivative.

       S = aL + a^2[sqrt(3) - 12]/2

   dS/da = L + 2a[sqrt(3) - 12]/2

   dS/da = L + a[sqrt(3) - 12]

Now, we can set the slope to 0 and find an equation that
represents the point of maximum surface area

             dS/da = L + a[sqrt(3) - 12]

                 0 = L + a[sqrt(3) - 12]
   
   a[sqrt(3) - 12] = -L

                 a = -L/[sqrt(3) - 12]

                 a = L/[12 - sqrt(3)]

This value of a is the value at the maximum surface area.
If we put that into our equation for surface area we get
the value of the maximum surface area:

   S = aL + a^2[sqrt(3) - 12]/2   <===   a = L/[12 - sqrt(3)]

Substitute that and solve for S.  It will equal the maximum
value you were given.

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


Date: 07/16/2002 at 04:48:22
From: Peter Koloszar
Subject: Thank you (Calculus-max. and min.)

Thanks a million for your extensive and prompt reply. I am 
really pleased and you guys are wonderful. Thank you very 
much. 

Peter Koloszar.
Associated Topics:
College Calculus
College Polyhedra
High School Calculus
High School Polyhedra

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/