I have no clue how to do this problem. please help.
A rectangle has one side on the x -axis and two vertices on the curve:
4/(2+x^2)
Find the vertices of the rectangle with maximum area.
Let the rectangle be symmetric. It has vertices at (-a, 0) and (a, 0). Then the width of the rectangle is 2a and the height of the rectangle is $\displaystyle \frac{4}{2 + a^2}$.
Then the area of the rectangle is $\displaystyle A = 2a \left(\frac{4}{2 + a^2} \right) = \frac{8a}{2 + a^2}$.
Use calculus to find the value of a that maximises A.