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

Answers (1)

Let assume that length and breadth of rectangle inscribed in a circle is l and b respectively
and the radius of the circle is r

Now, by Pythagoras theorem
a = \sqrt{l^2+b^2}\\
a = 2r
4r^2 = l^2+b^2\\ l = \sqrt{4r^2 - b^2}
Now, area of reactangle(A) = l \times b
A(b) = b(\sqrt{4r^2-b^2})
A^{'}(b) = \sqrt{4r^2-b^2}+b.\frac{(-2b)}{2\sqrt{4r^2-b^2}}\\ = \frac{4r^2-b^2-b^2}{\sqrt{4r^2-b^2}} = \frac{4r^2-2b^2}{\sqrt{4r^2-b^2}}
A^{'}(b) = 0 \\ \frac{4r^2-2b^2}{\sqrt{4r^2-b^2}} = 0\\ 4r^2 = 2b^2\\ b = \sqrt2r
Now,
A^{''}(b) = \frac{-4b(\sqrt{4r^2-b^2})-(4r^2-2b^2).\left ( \frac{-1}{2(4r^2-b^2)^\frac{3}{2}}.(-2b) \right )}{(\sqrt{4r^2-b^2})^2}\\ A^{''}(\sqrt2r) = \frac{(-4b)\times\sqrt2r}{(\sqrt2r)^2} = \frac{-2\sqrt2b}{r}< 0
Hence,  b = \sqrt2r is the point of maxima
l = \sqrt{4r^2-b^2}=\sqrt{4r^2-2r^2}= \sqrt2r
Since, l = b we can say that the given rectangle is a square
Hence, of all the rectangles inscribed in a given fixed circle, the square has the maximum area

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