Mathematics - Calculus

Suppose that the vertex of the rectangle in the first quadrant is [c, (1/2)√(1 - c2)], where c > 0, then:
Vertex in second quadrant: [-c, (1/2)√(1 - c2)]
Vertex in third quadrant: [-c, (-1/2)√(1 - c2)]
Vertex in fourth quadrant: [c, (-1/2)√(1 - c2)]
So the width and length of the rectangle formed are 2c and √(1 - c2)
Thus, area is:
A = 2c√(1 - c2)
Taking differentiation of A w.r.t. c:
dA/dc = 2√(1 - c2) + 2[c/√(1 - c2)] d(1 - c2)/dc
= 2√(1 - c2) - 4c2/√(1 - c2)
So for dA/dc = 0:
2√(1 - c2) = 4c2/√(1 - c2)
2 - 2c2 = 4c2
c = 1/√3
So the dimentions are 2/√3 and √(2/3) with area = (2√2)/3

2009-08-21 20:11:35 補充：
Since d(1 - c^2)/dc = -2c,
dA/dc = 2√(1 - c^2) + 2[c/√(1 - c^2)] d(1 - c^2)/dc
= 2√(1 - c^2) + 2[c/√(1 - c^2)](-2c)
= 2√(1 - c^2) - 4c^2/√(1 - c^2)

2009-08-21 20:12:04 補充：
Since d(1 - c^2)/dc = -2c,
dA/dc = 2√(1 - c^2) + 2[c/√(1 - c^2)] d(1 - c^2)/dc
= 2√(1 - c^2) + 2[c/√(1 - c^2)](-2c)
= 2√(1 - c^2) - 4c^2/√(1 - c^2)

2009-08-21 21:23:44 補充：
Thanks for the correction and sorry for the mistake.
Pls. find the correct ans as follows:
http://i388.photobucket.com/albums/oo325/loyitak1990/Aug09/Crazydiff6.jpg