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

1 Answer

  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
  • Perimeter=$2ab$
Step 1:
Let the length and breadth of the rectangle inscribed in a circle of radius 'a' be 'x' and 'y' respectively.
Differentiating with respect to x we get
Step 2:
On double differentiation we get
Step 3:
For $P(x)$ to be minimum $P'(x)=0$ and $P''(x)<0$
From (1),$P'(x)=0$
$\Rightarrow 2\bigg[1-\large\frac{x}{\sqrt{4a^2-x^2}}\bigg]$$=0$
$\Rightarrow 1-\large\frac{x}{\sqrt{4a^2-x^2}}$$=0$
$\Rightarrow 1=\large\frac{x}{\sqrt{4a^2-x^2}}$
$\Rightarrow 4a^2-x^2=x^2$
$x=\pm \sqrt 2 a$
Taking only positive value $x=\sqrt 2a$
Step 4:
From equation (3)
$\qquad=\large\frac{-8a^2}{(2a^2)^{\Large\frac{3}{2}}}$ which is +ve
$\Rightarrow P(x)$ is maximum at $x=\sqrt 2a$
Step 5:
From (1)
$y=\sqrt 2a$
Thus $x=y$
Hence the rectangle is a square of side $\sqrt 2a$
answered Aug 8, 2013 by sreemathi.v
edited Aug 19, 2013 by sharmaaparna1

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