find the rectangle of greatest area that can be inscribed in the circle x^2+y^2=36
We can always set up a coordinate system so that the origin is at the center of the circle and the x-axis is parallel to one side of the rectangle. In that coordinate system, the circle has equation $\displaystyle x^2+ y^2= R^2$ for some R (the radius of the circle). If we call the corner of the square in the first quadrant $\displaystyle (x_0, y_0)$, then, because the vertex lies oh the circle, $\displaystyle x_0^2+ y_0^2= R^2$ so $\displaystyle y_0= \sqrt{R^2- x_0^2$.
The lengths of the sides of the rectangle are, by symmetry, just twice those x and y values so the area of the rectangle is $\displaystyle (2x_0)(2y_0)= 4x_0\sqrt{R^2- x_0^2$. Replace $\displaystyle x_0$ by x and find the value of x that maximizes $\displaystyle A= 4x(R^2- x^2)^{1/2}$.