Strongest Possible Beam From a Log

Date: 07/11/2002 at 22:37:41
From: Bill Gemberling
Subject: Calculus - Optimization Rectangular Beam

The strength of a rectangular beam is directly proportional to the 
product of its width and the square of the depth of a cross section.  
Find the dimensions of the strongest beam that can be cut from a 
cylindrical log of radius 'a'.

There is a drawing included with the problem in the book that makes 
the corners of the rectangle come to the edge of the log. I know I'm 
missing some formula to figure this out.  I understand the product to 
be:
     wd^2 = S,

where w = width, d = depth, S = strength of a rectangular beam.

Thank you for your help.

Bill Gemberling
billgember@yahoo.com

Date: 07/13/2002 at 12:58:07
From: Doctor Fwg
Subject: Re: Calculus - Optimization Rectangular Beam

Dear Bill,

Here is one way to treat this problem:

  If the width and depth of a rectangular beam are W and D, 
  respectively, and the strength of the beam (S), in terms of 
  W and D, is

    S = kWD^2

  where k is a constant of proportionality, find the dimensions of 
  the strongest rectangular beam that can be cut from a uniform
  cylindrical log having a radius of value A.

Start with:

  1)  S = kWD^2.

But if the distance between the center of the cut beam and one corner 
is A, and the horizontal distance between the center of the cut beam 
and one side is X, and the vertical distance between the center of 
the cut beam and its top side is Y, then, let W = 2X and D = 2Y.

This changes S to:

  2)  S = k(2X)(2Y)^2 

        = 8k(X)(Y)^2.

These stipulations allow the following:

  3)  X^2 + Y^2  =  A^2.

So:

  4)  Y^2 = A^2 - X^2

Now using the right hand side (RHS) of Equation 4 in Equation 2
(to eliminate Y^2), one gets:

  5)  S = 8k(X)(A^2 - X^2) 

       = 8k(A^2X - X^3).

Now, if one takes the first derivative of S with respect to (WRT) X 
and sets that result to zero, one can find the value of X (in terms 
of A, which is a constant) which causes S to be a maximum (because 
the second derivative of S WRT X is negative).

One should get the result that 

  X = [Sqrt(1/3)]A 

and 

  Y = [Sqrt(2/3)]A.

Remember that W = 2X and D = 2Y.

It will also be possible to show that 

  Y/X = D/W = sqrt(2).  

I hope this helps.

With Best Regards, 

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

Date: 07/14/2002 at 15:54:40
From: Bill Gemberling
Subject: Thank you (Calculus - Optimization Rectangular Beam)

Doctor Fwg,

Thank you for your help with the Optimization Rectangular 
Beam problem.  I am taking Calculus for the first time and 
this has been helpful.  Once again thank you for your help.

Bill Gemberling