Question

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

Answer 1

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.