So let a and b the abscissas the base :
1 A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=1-x2. What are the dimensions of such a rectangle with the greatest possible area?
The upper corners points C and D are on the parabola.
| a b
Consider that C has an abscissa of b and D an abscissa of a (since it's a rectangle, with a base on the x-axis, they have the same abscissa)
C has coordinates (b,1-b²) and D : (a,1-a²) (remember that C and D are on the parabola)
But since it's a rectangle, CD is parallel to AB, and thus parallel to the x-axis. Hence C and D have the same ordinate
So in fact D has coordinates (a,1-b²)
---> a²-b²=0 ---> (a-b)(a+b)=0
So either a=b, which would make an area of 0, either a=-b.
Since b is the greatest, b is positive.
The area is given by
This is the function to maximize.
Remember that b is between 0 and the positive y-intercept of the parabola (1-x²=0 --> x=1) (because the rectangle is inscribed and because b>0)