Let a rectangle of length l and breadth b be inscribed in the given circle of radius a.
Then,the diagonal passes through the centre and is of length \(2a \space cm.\)
Now,by applying the Pythagoras theorem,we have:
\((2a)^2=l^2+b^2\)
\(⇒b^2=4a^2-l^2\)
\(⇒b=\sqrt{4a^2-l^2}\)
⧠Area of the rectangle\(,A=l\sqrt{4a^2-l^2}\)
\(∴\frac{dA}{dl}\)\(=\sqrt{4a^2-l^2}+l\frac{1}{2√4a^2-l^2}(-2l)=\sqrt{4a^2-l^2}-\frac{l^2}{\sqrt{4a^2-l^2}}\)
\(=\frac{4a^2-2l^2}{\sqrt{4a^2-l^2}}\)
\(\frac{d^2A}{dl^2}=\frac{\sqrt{4a^2-l^2}(-4l)-(4a^2-2l^2)\frac{(-2l)}{2\sqrt{4a^2-l^2}}}{(4a^2-l^2)}\)
\(=\frac{(4a62-l62)(-4l)+l(4a^2-2l^2)}{(4a^2-l^2)\frac{3}{2}}\)
\(=\frac{-12a^2l+2l^3}{(4a^2-l^2)\frac{3}{2}}\)=\(\frac{-2l(6a^2-l^2)}{(4a^2-l^2)\frac{3}{2}}\)
Now\(,\frac{dA}{dl}=0\) gives \(4a^2=2l^2⇒l=\sqrt{2}a\)
\(⇒b=\sqrt{4a^2-2a^2}=\sqrt{2a^2}=\sqrt{2}a\)
Now,then\( l=\sqrt{2}a\)
\(\frac{d^2A}{dl^2}\)=\(\frac{-2(\sqrt{2}a)(6a^2-2a^2)}{2\sqrt{2}a^3}\)=\(\frac{-8\sqrt{2}a^3}{2\sqrt{2}a^3}=-4<0\)
∴By the second derivative test,when\( l=\sqrt{2}a\),then the area of the rectangle is the
maximum.
Since \(l=b=\sqrt{2}a\),the rectangle is a square.
Hence,it has been proved that of all the rectangles inscribed in the given fixed circle,
the square has the maximum area.
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