Find the area of the largest rectangle that can be inscribed in the ellipse x2a2 + y2b2 = 1.

The area of a rectangle inscribed in an ellipse with semi-major axis a and semi-minor axis b is given by the formula A = 4xy, where x and y are the x and y coordinates of a point on the ellipse.

The equation of the ellipse is given by (x^2/a^2) + (y^2/b^2) = 1. We can solve this equation for y to get y = b*sqrt(1 - x^2/a^2).

Substituting this into the area formula gives A = 4xbsqrt(1 - x^2/a^2).

To find the maximum area, we need to find the maximum of this function. This can be done by taking the derivative of the function with respect to x and setting it equal to zero.

The derivative of A with respect to x is A' = 4bsqrt(1 - x^2/a^2) - 4bx^2/(a^2sqrt(1 - x^2/a^2)).

Setting this equal to zero and solving for x gives x = a/sqrt(2).

Substituting this back into the equation for y gives y = b/sqrt(2).

Substituting these values into the area formula gives A = 4ab/(2) = 2ab.

So, the maximum area of a rectangle that can be inscribed in the ellipse is 2ab.