Question

14. Find the maximum area of any rectangle which may be inscribed in a circle of radius 1. (Area=2)​

Answer 1

 There are two ways to solve this problem.

 

 Here is the answer using geometry. (Attachment included)

 

 A rectangle gives two coincidental right triangles when divided along its diagonal.

 So let the right triangle be inscribed in a semicircle, which diameter is its hypotenuse.

 The maximum area is when the height becomes the radius. The inscribed triangle has a maximum area of 1.

 The coincidental triangle has the same area, so two triangles on a semicircle(a square) have a maximum area of 2.

 

 Here is the answer using algebra.

 

 Let the circle be a unit circle. The equation is $\sf{x^2+y^2=1}$.

 Let the length and width be $l$ and $w$.

 Since the rectangle's diagonal is the diameter, we have $\sf{l^2+w^2=4}$.

$\therefore \sf{w^2=4-l^2}$

 The area of the rectangle is $wl$. The squared area is $\sf{(wl)^2}$.

 The squared area can be calculated.

$\sf{w^2l^2=(4-l^2)l^2}$

$\sf{=-l^4+4l^2}$

$\sf{=-(l^2-2)^2+4\:(wl>0)}$

$\therefore \sf{wl=\sqrt{-(l^2-2)^2+4} }$

 Hence the maximum area is 2.

 When the square with a side of √2 is inscribed, the maximum area is 2.

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