Show that the rectangle of max perimeter which can be inscribed in a circle of radius 'a' is a - Maths - Application of Derivatives

Question

Show that the rectangle of max perimeter which can be inscribed
in a circle of radius 'a' is a square of side
a√2

Aakash Sharma · Accepted Answer

Let ABCD be a rectangle in a given circle of radius a with centre at O Let AB = 2x and AD = 2y be the sides of the rectangle Then, AM2 + OM2 = OA2 [ $∵$ Î” OAM is a right angles triangle] â‡’ x2 + y2 = a2 $\Rightarrow y = \sqrt{a^{2} - x^{2}}$ Let P be the perimeter of the rectangle ABCD Then, P = 4x + 4y $\Rightarrow P = 4 x + 4 \sqrt{a^{2} - x^{2}} [Using (i)] \Rightarrow \frac{d P}{d x} = 4 - \frac{4 x}{\sqrt{a^{2} - x^{2}}}$

For maximum or minimum, values of P, we have $\frac{d P}{d x} = 0 \Rightarrow 4 - \frac{4 x}{\sqrt{a^{2} - x^{2}}} = 0 \Rightarrow 4 = \frac{4 x}{\sqrt{a^{2} - x^{2}}} \Rightarrow \sqrt{a^{2} - x^{2}} = x \Rightarrow a^{2} - x^{2} = x^{2} \Rightarrow 2 x^{2} = a^{2} \Rightarrow x = \frac{a}{\sqrt{2}}$ Now, $\frac{d^{2} P}{d x^{2}} = \frac{- 4 {\sqrt{a^{2} - x^{2}} 1 - \frac{x (- x)}{\sqrt{a^{2} - x^{2}}}}}{(\sqrt{a^{2} - x^{2}})^{2}} = \frac{- 4 a^{2}}{(a^{2} - x^{2})^{\frac{3}{2}}} ∴ (\frac{d^{2} P}{d x^{2}})_{x = \frac{a}{\sqrt{2}}} = \frac{- 4 a^{2}}{(a^{2} - \frac{a^{2}}{2})^{\frac{3}{2}}} = \frac{- 8 \sqrt{2}}{a} < 0$ Thus, P is maximum when $x = \frac{a}{\sqrt{2}}$ Putting $x = \frac{a}{\sqrt{2}}$ in (1), we obtain $y = \frac{a}{\sqrt{2}}$ Therefore, $x = y = a \sqrt{2} \Rightarrow 2 x = 2 y$ Hence, P is maximum when the rectangle is square of side $2 x = \frac{2 a}{\sqrt{2}} = \sqrt{2} a$

Show that the rectangle of max. perimeter which can be inscribed in a circle of radius 'a' is a square of side a√2.