Apply of Differentiation (急!!)

Suppose that the radius of sector is x cm, then the arc length will be (k - 2x) cm.
Moreover, the angle subtended at the centre of the sector is:
θ = (k - 2x)/x = (k/x - 2) radians
Applying the formula of sector area:
A = x2θ/2
= x2(k/x - 2)/2
= (kx/2 - x2) cm2
Taking differentiation of A w.r.t. x:
dA/dx = (k/2 - 2x)
d2A/dx2 = - 2
When dA/dx = 0, x = k/4 and since d2A/dx2 < 0, it is for sure that x = k/4 will give a maximum of A.
Hence the max. area is:
Amax = [k(k/4)/2 - (k/4)2] = k2/16 cm2

Let radius of the sector = r and angle of sector = x.
k = rx + 2r = r( x + 2)......... (1) ( x in radian)
Area, A = r^2x/2.......(2)
From (1) 0 = ( x + 2) + r dx/dr, so dx/dr = -(x + 2)/r.
From (2)
dA/dr = [r^2 dx/dr + 2rx]/2
= [- r^2(x + 2)/r + 2rx]/2
= [-(x + 2)r + 2rx]/2
= (- 2r + rx)/2.
Put dA/dr = 0, we get
2r = rx
x = 2.
when x = 2, from (1), r = k/4.
so A max. = (k/4)^2(2/2) = (k/4)^2.