Show that of all the rectangles inscribed in a given fixed circle,the square has the maximum area.

(N/A) Let a rectangle of length $l$ and breadth $b$ be inscribed in the given circle of radius $R$.
The diagonal of the rectangle passes through the center of the circle and is equal to the diameter,$2R$.
By applying the Pythagoras theorem,we have:
$(2R)^{2} = l^{2} + b^{2}$
$\Rightarrow b^{2} = 4R^{2} - l^{2}$
$\Rightarrow b = \sqrt{4R^{2} - l^{2}}$
Area of the rectangle,$A = l \times b = l \sqrt{4R^{2} - l^{2}}$
To maximize the area,we maximize $A^{2}$. Let $S = A^{2} = l^{2}(4R^{2} - l^{2}) = 4R^{2}l^{2} - l^{4}$.
Differentiating with respect to $l$:
$\frac{dS}{dl} = 8R^{2}l - 4l^{3}$
Setting $\frac{dS}{dl} = 0$:
$4l(2R^{2} - l^{2}) = 0$
Since $l \neq 0$,we have $l^{2} = 2R^{2} \Rightarrow l = R\sqrt{2}$.
Now,$b = \sqrt{4R^{2} - (R\sqrt{2})^{2}} = \sqrt{4R^{2} - 2R^{2}} = \sqrt{2R^{2}} = R\sqrt{2}$.
Since $l = b = R\sqrt{2}$,the rectangle is a square.
Checking the second derivative:
$\frac{d^{2}S}{dl^{2}} = 8R^{2} - 12l^{2}$
At $l^{2} = 2R^{2}$,$\frac{d^{2}S}{dl^{2}} = 8R^{2} - 12(2R^{2}) = 8R^{2} - 24R^{2} = -16R^{2} < 0$.
Since the second derivative is negative,the area is maximum when $l = b = R\sqrt{2}$.
Thus,it is proved that of all rectangles inscribed in a fixed circle,the square has the maximum area.