Lesson Maximize the area of a trapezoid

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Maximize the area of a trapezoid


Problem 1

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?

Solution 

It is clear, that under given conditions, the trapezoid is isosceles with the shorter base of 10 cm long
and lateral sides of 10 cm long, each.


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

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

   <- x ->


The mid-line is  %2810%2B%2810%2B2x%29%29%2F2 = 10+x  cm long.

The height is  sqrt%2810%5E2-x%5E2%29.


The area of the trapezoid  is   A(x) = %2810%2Bx%29%2Asqrt%28100-x%5E2%29.


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

    %28dA%29%2F%28dx%29 = sqrt%28100-x%5E2%29 - %281%2F2%29%2A%28%2810%2Bx%29%2Fsqrt%28100-x%5E2%29%29%2A%28-2x%29

    %28dA%29%2F%28dx%29 = sqrt%28100-x%5E2%29 - %28%2810%2Bx%29%2Ax%29%2Fsqrt%28100-x%5E2%29


Equate it to zero

    sqrt%28100-x%5E2%29 - %28%2810%2Bx%29%2Ax%29%2Fsqrt%28100-x%5E2%29 = 0.


Simplify and find x

    sqrt%28100-x%5E2%29 = %28%2810%2Bx%29%2Ax%29%2Fsqrt%28100-x%5E2%29

    100-x%5E2%29 = (10+x)*x

    100-x%5E2 = 10x+%2B+x%5E2

    2x%5E2+%2B+10x+-+100 = 0

    x%5E2+%2B+5x+-+50 = 0

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


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


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


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