Question 1192493
.
Three sides of a trapezoid are each 10 cm long. 
How long is the 4th side, when the area of the trapezoid has the greatest value?
~~~~~~~~~~~~~~~~~


<pre>
It is clear, that under given conditions, the trapezoid is isosceles with the shorter base of 10 cm long.


See my sketch below.  The upper base is 10 cm;  the lower base is 10+2x cm long.

                  10
         +------------------+
        /|                   \
       / |                    \
  10  /  |                     \  10
     /   |                      \
    /    |                       \
   +------------------------------+

   <- x ->


The mid-line is  {{{(10+(10+2x))/2}}} = 10+x  cm long.

The height is  {{{sqrt(10^2-x^2)}}}.


The area of the trapezoid  is   A(x) = {{{(1+x)*sqrt(100-x^2)}}}.


To find the maximum of A(x), take the derivative

    {{{(dA)/(dx)}}} = {{{sqrt(100-x^2)}}} - {{{(1/2)*((10+x)/sqrt(100-x^2))*(-2x)}}}

    {{{(dA)/(dx)}}} = {{{sqrt(100-x^2)}}} - {{{((10+x)*x)/sqrt(100-x^2)}}}


Equate it to zero

    {{{sqrt(100-x^2)}}} - {{{((10+x)*x)/sqrt(100-x^2)}}} = 0.


Simplify and find x

    {{{sqrt(100-x^2)}}} = {{{((10+x)*x)/sqrt(100-x^2)}}}

    {{{100-x^2)}}} = (10+x)*x

    {{{100-x^2}}} = {{{10x + x^2}}}

    {{{2x^2 + 10x - 100}}} = 0

    {{{x^2 + 5x - 50}}} = 0

    (x-5)*(x+10) = 0


Of two roots,  x= 5 and x= -10, take the positive one, which gives the maximum to the trapezoid's area.


<U>ANSWER</U>.  The area of trapezoid is maximum when the 4-th side is 10+2*5 = 20 cm long.
</pre>

Solved.