Question

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

Answer 1

`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)` with `0< a,b < 1`. The coordinates of the others are `P(+-a,+-b)` with the constraint:

`a^2+b^2 = 1`

so that:

`b= sqrt(1-a^2)`

and the perimeter is:

`P= 2a+2b = 2(a+sqrt(1-a^2))`

Evaluate the first derivative:

`(dP)/(da) = 2-(2a)/sqrt(1-a^2)`

And identify critical points solving the equation:

`(dP)/(da) = 0`

`1= a/sqrt(1-a^2)`

`a= sqrt(1-a^2)`

`a^2 = 1-a^2`

`a=1/sqrt2`

Evaluate the second derivative:

`(d^2P)/(da^2) = (-2sqrt(1-a^2)-2a/sqrt(1-a^2))/(1-a^2)`

`(d^2P)/(da^2) = (-2(1-a^2)-2a)/((1-a^2)sqrt(1-a^2))`

`(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=1/sqrt2` and the rectangle is a square.