Maximize Area of Trapezoid

Date: 07/28/2003 at 09:35:34
From: Jupe
Subject: Find d so that the area is maximum

Given an isosceles trapezoid with three of its sides of length 10 cm:
  
                       d cm
             +------------------------+
              \                      /
         10 cm \                    / 10 cm
                \                  /
                 \                /
                  +--------------+
                       10 cm

find d so that the area is maximum.

Date: 07/29/2003 at 22:29:48
From: Doctor Jeremiah
Subject: Re: Find d so that the area is maximum

Hi Jupe,

Let's break the trapezoid into pieces so that we can more easily 
calculate the area:

                x        10        x
             +----+--------------+----+
              \   |              |   /
            10 \  | h          h |  / 10
                \ |              | /
                 \|              |/
                  +--------------+
                         10

The area of this shape is the sum of the areas of the two triangles 
and the rectangle. The height can be found using the Pythagorean 
theorem:

  x^2 + h^2 = 10^2

        h^2 = 10^2 - x^2

          h = sqrt(10^2 - x^2)

The area of one triangle is:  hx/2

The area of the rectangle is: 10h

So the total area is:         2(hx/2) + 10h = hx + 10h = (x+10)h

We have a value for h so if we plug that into the area:

  area = (x+10) sqrt(10^2 - x^2)

  area = sqrt(10^2 (x^2+20x+100) - x^2 (x^2+20x+100))

  area = sqrt(-x^4 - 20x^3 + 2000x + 10000)

To find the maximum area we need the place where this slope becomes 0.  
We can either do that graphically or with a derivative:

  d/dx( area ) = d/dx( sqrt(-x^4 - 20x^3 + 2000x + 10000) )

             0 = 1/2 d/dx( -x^4 - 20x^3 + 2000x + 10000 )
                 ----------------------------------------
                    sqrt(-x^4 - 20x^3 + 2000x + 10000)

Which after some rearrangement and simplification gives:

             4 x^3 + 60 x^2 = 2000

And that can be solved as:

             x = 5

To we check our answer, we can plug into the area equation the number 
5, a number slightly more than 5, and a number slightly less than 5:

  area = sqrt(-x^4 - 20x^3 + 2000x + 10000)

  when x=5.01   area = sqrt(16874.95496)

  when x=5      area = sqrt(16875)

  when x=4.99   area = sqrt(16874.95504)

Notice that the value at 5 is larger than the others around it.

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

Date: 07/30/2003 at 09:52:41
From: Jupe
Subject: Find d so that the area is maximum

Is the only method involved when you're asked to find maximum and 
minimum the differentiation method? And is there another method to do 
the same question other than this and the maximum of angle x?

Date: 07/30/2003 at 18:00:19
From: Doctor Jeremiah
Subject: Re: Find d so that the area is maximum

Hi Jupe,

As I mentioned, you can graph the equation for area and physically
look at the graph to find a zero slope, which will tell you where the
maximum value is.

That's what a derivative does - it tells you the slope of the graph,
and when you set the derivative to zero it tells you where the zero
slope is. So basically doing a derivative is the same thing as looking
at the graph yourself.

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