greatest area inscribed in a circle

**antikv** · April 11th 2011, 02:03 PM

find the rectangle of greatest area that can be inscribed in the circle x^2+y^2=36

**e^(i*pi)** · April 11th 2011, 02:26 PM

The diagonal of the rectangle will be the diameter and I believe the shape is a square

**HallsofIvy** · April 12th 2011, 10:09 AM

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 $x^2+ y^2= R^2$ for some R (the radius of the circle). If we call the corner of the square in the first quadrant $(x_0, y_0)$ , then, because the vertex lies oh the circle, $x_0^2+ y_0^2= R^2$ so $y_0= \sqrt{R^2- x_0^2$ .

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 $(2x_0)(2y_0)= 4x_0\sqrt{R^2- x_0^2$ . Replace $x_0$ by x and find the value of x that maximizes $A= 4x(R^2- x^2)^{1/2}$ .

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