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

May 25, 2018

$P = 2 \sqrt{2}$

#### 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 $P \left(a , b\right)$ with $0 < a , b < 1$. The coordinates of the others are $P \left(\pm a , \pm b\right)$ with the constraint:

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

so that:

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

and the perimeter is:

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

Evaluate the first derivative:

$\frac{\mathrm{dP}}{\mathrm{da}} = 2 - \frac{2 a}{\sqrt{1 - {a}^{2}}}$

And identify critical points solving the equation:

$\frac{\mathrm{dP}}{\mathrm{da}} = 0$

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

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

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

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

Evaluate the second derivative:

$\frac{{d}^{2} P}{{\mathrm{da}}^{2}} = \frac{- 2 \sqrt{1 - {a}^{2}} - 2 \frac{a}{\sqrt{1 - {a}^{2}}}}{1 - {a}^{2}}$

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

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

so that the critical point is a local maximum.

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