SOLUTION: Two verticies of a rectangle are on the x-axis and the other two vertices on the curve y=8/(x^2+4). Find the maximum area of the rectangle. I did find y'=-16x/(x^2+4)^2. I could

The rectangle has the horizontal side 2x and vertical side  .


So the area of the rectangle is   


        A(x) =  = .    (1)


To find the maximum of the area, we should calculate the derivative  A'(x)  and equate it to zero.



It is enough to calculate the NUMERATOR of the derivative: it is


    16*(x^2+1) - 16x*(2x) = 16x^2 + 16 - 32x^2 = 16 - 16x^2.



After that, we equate the derivative to zero - - - again, it is enough to equate the NUMERATOR to zero, which gives


    16 = 16x^2,  or  x^2 = 1,  or  x = +/-1, of which we select  x= 1.



So the maximum area of the rectangle is  (use formula (1)  at x= 1 )    


          =  =  = 8.    ANSWER