addtional maths微分問題(好急,要快)

圓的方程為
x^2 + y^2 = a^2

let P (x,y) be a point on the circle, x > 0
and P is the corner of the rectangle

Area = A = 2xy
= 2 x sqrt (a^2 – x^2)

dA / dx = 2 * [sqrt (a^2 – x^2) + x * (-2x) * (1/2) / sqrt (a^2 – x^2) ]

put dA / dx = 0
and on solving, x = a / sqrt 2

hence y = a / sqrt 2

area = 2 xy = a^2

2008-05-12 20:19:27 補充：
**** let P (x,y) be a point on the circle, x ＞ 0

仲有
點解我唔 verify 佢係 max 定 min 呢
其實呢次可以唔駛
因為 min area 係 0 (出現係 x = 0 or x = a)

According to Brahmagupta's formula, the area is maximum when all the side length of the quadrilateral are the same, i.e. this quadrilateral is a square.

Let the the length of the square be x

x^2 = a^2 + a^2

x=sq. root(2)*a