A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=9–x2. What are the dimensions of such a rectangle with the greatest possible area?

Answer 1

Consider half of the rectangle which lies completely in the first quadrant.
The lower left corner is the origin (0,0), and the upper right corner lies on the curve y=9-x².

The area of the (half) rectangle is therefore:

A(x)
=x*y
=x*(9-x²)
=9x-x³

Differentiate with respect to x,
and equate A'(x) to zero to get the value of x that will result in the maximum area. The area of the required rectangle is twice A(x) because the other half of the rectangle is in the second quadrant.

Do check that A"(x)<0 to confirm that the area is a maximum (and not a minimum).