How do you maximize the perimeter of a rectangle inside a circle with equation: $x^{2} + y^{2} = 1$ $x^2+y^2=1$ ?

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How do you maximize the perimeter of a rectangle inside a circle with equation: $x^{2} + y^{2} = 1$ $x^2+y^2=1$ ?

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Answer 1 · 2018-05-25T11:54:17

$P = 2 \sqrt{2}$ $P=2sqrt2$

Explanation:

For symmetry reasons we can assume the rectangle has sides parallel to the axes. In such case if the corner in the first quadrant is the point $P (a, b)$ $P(a,b)$ with $0 < a, b < 1$ $0< a,b < 1$ . The coordinates of the others are $P (\pm a, \pm b)$ $P(+-a,+-b)$ with the constraint:

$a^{2} + b^{2} = 1$ $a^2+b^2 = 1$

so that:

$b = \sqrt{1 - a^{2}}$ $b= sqrt(1-a^2)$

and the perimeter is:

$P = 2 a + 2 b = 2 (a + \sqrt{1 - a^{2}})$ $P= 2a+2b = 2(a+sqrt(1-a^2))$

Evaluate the first derivative:

$\frac{d P}{d a} = 2 - \frac{2 a}{\sqrt{1 - a^{2}}}$ $(dP)/(da) = 2-(2a)/sqrt(1-a^2)$

And identify critical points solving the equation:

$\frac{d P}{d a} = 0$ $(dP)/(da) = 0$

$1 = \frac{a}{\sqrt{1 - a^{2}}}$ $1= a/sqrt(1-a^2)$

$a = \sqrt{1 - a^{2}}$ $a= sqrt(1-a^2)$

$a^{2} = 1 - a^{2}$ $a^2 = 1-a^2$

$a = \frac{1}{\sqrt{2}}$ $a=1/sqrt2$

Evaluate the second derivative:

$\frac{d^{2} P}{d a^{2}} = \frac{- 2 \sqrt{1 - a^{2}} - 2 \frac{a}{\sqrt{1 - a^{2}}}}{1 - a^{2}}$ $(d^2P)/(da^2) = (-2sqrt(1-a^2)-2a/sqrt(1-a^2))/(1-a^2)$

$\frac{d^{2} P}{d a^{2}} = \frac{- 2 (1 - a^{2}) - 2 a}{(1 - a^{2}) \sqrt{1 - a^{2}}}$ $(d^2P)/(da^2) = (-2(1-a^2)-2a)/((1-a^2)sqrt(1-a^2))$

$\frac{d^{2} P}{d a^{2}} = \frac{- 2 - 2 a^{2} - 2 a}{(1 - a^{2}) \sqrt{1 - a^{2}}} < 0$ $(d^2P)/(da^2) = (-2-2a^2-2a)/((1-a^2)sqrt(1-a^2)) <0$

so that the critical point is a local maximum.

Then the perimeter is maximum when $a = b = \frac{1}{\sqrt{2}}$ $a=b=1/sqrt2$ and the rectangle is a square.

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How do you maximize the perimeter of a rectangle inside a circle with equation: $x^{2} + y^{2} = 1$ $x^2+y^2=1$ ?

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Explanation:

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How do you maximize the perimeter of a rectangle inside a circle with equation: $x^{2} + y^{2} = 1$ $x^2+y^2=1$ ?