Show that of all the rectangle inscribed in a given circle,the square has the max area
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.
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
To show that the square has the maximum area among all the rectangles inscribed in a given circle, we need to compare the area of the square with that of any other rectangle.
Let's consider a rectangle inscribed in a circle. The two opposite vertices of the rectangle lie on the diameter of the circle. Let's call the length of the rectangle's shorter side "x" and the length of the longer side "y". Since the rectangle is inscribed in the circle, the diagonal of the rectangle is equal to the diameter of the circle.
To compare the area of the square with the area of this rectangle, we need to express the area of the rectangle in terms of x and y.
The diagonal of the rectangle is given by the Pythagorean theorem as:
diagonal^2 = x^2 + y^2
Since the diagonal is equal to the diameter of the circle, we can substitute it with 2r, where r is the radius of the circle:
(2r)^2 = x^2 + y^2
4r^2 = x^2 + y^2 ----(1)
The area of the rectangle is given by:
Area = x * y
To determine the maximum area, we can use calculus. By applying the AM-GM (Arithmetic Mean-Geometric Mean) inequality, we know that the arithmetic mean of two numbers is always greater than or equal to the geometric mean of those numbers when both numbers are positive.
For x^2 and y^2, their arithmetic mean is (x^2 + y^2)/2, and their geometric mean is sqrt(x^2 * y^2) = xy.
By the AM-GM inequality:
(x^2 + y^2)/2 ≥ sqrt(x^2 * y^2)
(x^2 + y^2) ≥ 2xy
Using equation (1), we can substitute x^2 + y^2 with 4r^2:
4r^2 ≥ 2xy
2r^2 ≥ xy
This inequality tells us that xy (the area of any rectangle inscribed in the circle) is less than or equal to 2r^2.
Now, if we consider a square inscribed in the same circle, all sides of the square would have length 2r (i.e., twice the radius of the circle). The area of the square is given by:
Area = (2r)^2 = 4r^2
Comparing the area of the square (4r^2) with the area of the rectangle (xy ≤ 2r^2), we see that 4r^2 is always greater than or equal to 2r^2. Therefore, the square has the maximum area among all the rectangles inscribed in a given circle.