Hello,
1 A rectangle is inscribed with its base on the xaxis and its upper corners on the parabola y=1x2. What are the dimensions of such a rectangle with the greatest possible area?
So let a and b the abscissas the base :
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 a b

The upper corners points C and D are on the parabola.
Consider that C has an abscissa of b and D an abscissa of a (since it's a rectangle, with a base on the xaxis, they have the same abscissa)
b>a
C has coordinates (b,1b²) and D : (a,1a²) (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 xaxis. Hence C and D have the same ordinate
So in fact D has coordinates (a,1b²)
Hence 1b²=1a²
> a²b²=0 > (ab)(a+b)=0
So either a=b, which would make an area of 0, either a=b.
> a=b.
Since b is the greatest, b is positive.
The area is given by $\displaystyle AB \times BC$
Distance AB=(ba)=2b
Distance BC=1b²
So $\displaystyle A(b)=2b(1b^2)$
This is the function to maximize.
Remember that b is between 0 and the positive yintercept of the parabola (1x²=0 > x=1) (because the rectangle is inscribed and because b>0)