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

1 Answer
May 25, 2018

#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.