urgent math help needed on max and min

Quote:

Originally Posted by jojoferni244

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 $\frac{4}{2 + a^2}$ .

Then the area of the rectangle is $A = 2a \left(\frac{4}{2 + a^2} \right) = \frac{8a}{2 + a^2}$ .

Use calculus to find the value of a that maximises A.