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# Apply of Differentiation (急!!)

A piece of wire of length k cm is bent to from a sector. Find the maximum area of the sector.

### 2 Answers

- 魏王將張遼Lv 71 decade agoFavorite Answer
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

Source(s): My Maths knowledge - wyLv 71 decade ago
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.