Question

Find the maximum value of Sin(A)Sin(B)Sin(C) if A, B, C are the angles of triangle.

(a) ^3√3⁄8

(b) ^3√4⁄8

(c) –^3√3⁄8

(d) ^π⁄8

The question was asked in a national level competition.

I would like to ask this question from Maxima and Minima of Two Variables in portion Maxima and Minima of Engineering Mathematics

Answer 1

The correct answer is (a) ^3√3⁄8

Best explanation: Given f(A,B,C)=Sin(A)Sin(B)Sin(c),

Since A, B, C are the angle of triangle, hence, C = 180 – (A+B),

hence, f(x,y) = Sin(x)Sin(y)Sin(x+y), where A = x and B = y

Hence, ^∂f⁄∂x = Cos(x)Sin(y)Sin(x+y) + Sin(x)Sin(y)Cos(x+y) = Sin(y)Sin(y+2x)

and, ^∂f⁄∂y = Sin(x)Cos(y)Sin(x+y) + Sin(x)Sin(y)Cos(x+y) = Sin(x)Sin(x+2y)

Hence, putting ^∂f⁄∂x and ^∂f⁄∂y = 0, we get (x,y)=(60,60), (120,120)

Hence, at (x,y) = (60,60)we get,r = -√3, s = -√3/2, t = -√3, hence, rt-s^2= ^9⁄4∂x>0

hence, r<0 andrt-s^2>0 hence, f(x,y) or f(A,B) have maximum value at (60,60)

Hence, at (x,y)=(120,120)we get,r=√3,s=√3/2,t=√3,hence,rt-s^2 = ^9⁄4∂x>0

And this value is ^3√3⁄8

hence, r>0 and rt-s^2 >0 hence, f(x,y) or f(A,B) have minimum value at (60,60)

and this value is –^3√3⁄8.