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

Find the maximum value of Sin(A)Sin(B)Sin(C) if A, B, C are the angles of triangle. Correct Answer 3√3⁄8

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-s2= 9⁄4∂x>0 hence, r<0 andrt-s2>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-s2 = 9⁄4∂x>0 And this value is 3√3⁄8 hence, r>0 and rt-s2 >0 hence, f(x,y) or f(A,B) have minimum value at (60,60) and this value is –3√3⁄8.

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