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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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