Show that of all the rectangle inscribed in a given circle,the square has the max area

Answer 1

If the sides of the rectangle are x and y, then since the diameter of the circle is the diagonal of the rectangle,

x^2+y^2 = d^2

The area is xy. The area squared

A = a^2 = x^2y^2 = x^2 (d^2-x^2)
= d^2 x^2 - x^4

a has a maximum where A does, so since

dA/dx = 2d^2 x - 4x^3 = 2x(d^2-2x^2)

A has a maximum where d^2 = 2x^2 = x^2+x^2

That is, where x=y.

Thus the rectangle is a square.

Answer 2

the the circle be x^2 + y^2 = r^2
then y = √(r^2 - x^2) or (r^2 - x^2)^(1/2)

let the length of the rectangle be 2x, and the height be 2y
then the vertex in quadrant I is (x, (r^2 - x^2)^(1/2) )

area = 2x(2y) = 4xy
= 4x((r^2 - x^2)^(1/2)), remember r is a constant
d(area)/dx
= 4x(1/2)((r^2 - x^2)^(1/2))^(-1/2) (-2x) + 4(r^2 - x^2)^(1/2)
= -4(r^2 - x^2)^(-1/2) [ x^2 - (r^2 - x^2 ]/(r^2 - x^2)
= -4 (2x^2 - r^2)/(r^2 - x^2)
= 0 for a max/min of area

2x^2 = r^2
√2 x = r

then in x^2 + y^2 = r^2
x^2 + y^2 = 2x^2
y^2 = x^2

thus x = y

the length is 2x, the width is 2y or 2x
the rectangle must be a square