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Maximum Area for Irrigation Channel


Date: 02/08/2002 at 23:35:06
From: Ingrid
Subject: Applied maximum and minimum problems in calculus

An irrigation channel made of concrete is to have a cross section in 
the form of an isosceles trapezoid, three of whose sides are 4 feet 
long. How should it be shaped to have a maximum area? Solve as a 
function of x.

In my class, we typically use two formulas to solve these kinds of 
problems, like surface area and volume. However, in this problem, I 
have no idea what to do after I know the area formula.

Ingrid


Date: 02/09/2002 at 01:56:35
From: Doctor Jeremiah
Subject: Re: Applied maximum and minimum problems in calculus

Hi Ingrid,

Any time someone says maximum or minimum you know that calculus is 
involved. Look at this graph:

                       |
                       |
                       |    +++++
                      +|      ^      +
                       |      |
                  +    |     max        +
                       |
   -------------+------+------------------+--
                       |
              +        |
                       |
            +          |
                       |

It has a maximum. But how can you tell? Well, the slope changes from 
positive to negative at the maximum. And to change from positive to
negative the slope must pass through zero. So the maximum of the graph 
is where the slope equals zero. And since the slope is just the 
derivative, the maximum is where the derivative equals zero.

So with an isosceles trapezoid we have a shape that looks like:

     +----------------------------------+
      \                                /
       \                              /
        \                            /
       4 \                          / 4
          \                        /
           \                      /
            \ a                a /
             +------------------+
                      4

Which is the same as:

     4 sin(x)         4          4 sin(x)
     +-------+------------------+-------+
      \      |                  |      /
       \     |                  |     /
        \ 4 cos(x)          4 cos(x) /
       4 \   |                  |   / 4
          \  |                  |  /
           \x|                  |x/
            \|                  |/
             +------------------+
                      4

Which if we rearrange the pieces we get:

     4 sin(x)         4
     +-------+------------------+
     |\      |                  |
     |x\     |                  |
     |  \ 4 cos(x)          4 cos(x)
     |   \   |                  |
 4 cos(x) \  |                  |
     |     \x|                  |
     |      \|                  |
     +-------+------------------+
     4 sin(x)         4


So the area is:

           A = ( 4+4sin(x) )( 4 cos(x) )
           A = 4( 4 cos(x) ) + 4sin(x)( 4 cos(x) )
           A = 16 cos(x) + 16 sin(x)cos(x)

Notice that sin(x)cos(x) is the same as sin(2x)/2

           A = 16 cos(x) + 16 sin(x)cos(x)
           A = 16 cos(x) + 16 sin(2x)/2
           A = 16 cos(x) + 8 sin(2x)

If we could just take the derivative of this, we would get the slope 
and we could set that to zero, find out what value of x makes the 
maximum area, and solve for the maximum area.

   d/dx( A ) = d/dx( 16 cos(x) + 8 sin(2x) )
       dA/dx = d/dx( 16 cos(x) ) + d/dx( 8 sin(2x) )
       dA/dx = 16 d/dx( cos(x) ) + 8 d/dx( sin(2x) )

First the first term:

       dA/dx = 16 d/dx( cos(x) ) + 8 d/dx( sin(2x) )
       dA/dx = 16 (-sin(x)) d/dx( x ) + 8 d/dx( sin(2x) )
       dA/dx = -16 sin(x) dx/dx + 8 d/dx( sin(2x) )
       dA/dx = -16 sin(x) + 8 d/dx( sin(2x) )

Now the next term:

       dA/dx = -16 sin(x) + 8 d/dx( sin(2x) )
       dA/dx = -16 sin(x) + 8 cos(2x) d/dx( 2x )
       dA/dx = -16 sin(x) + 8 cos(2x) 2 d/dx( x )
       dA/dx = -16 sin(x) + 16 cos(2x) d/dx( x )
       dA/dx = -16 sin(x) + 16 cos(2x) dx/dx
       dA/dx = -16 sin(x) + 16 cos(2x)

Remember, the maximum area is when the slope (derivative) is equal to 
zero:

           0 = -16 sin(x) + 16 cos(2x)
   16 sin(x) = 16 cos(2x)
      sin(x) = cos(2x)

There is only one good answer to this:  x=30

      sin(x) = cos(2x)
     sin(30) = cos(60)

So x=30 is the angle that makes the maximum area. That means the 
maximum area is:

        maxA = 16 cos(x) + 8 sin(2x)   <===   x=30
        maxA = 16 cos(30) + 8 sin(60)

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

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