已知A(0,a),B(0,b)是Y轴正正半轴上两点0,<a,<b,C(x,0)是X轴正半轴上任意一点,求当C在何位置时角ACB最大

已知A(0,a),B(0,b)是Y轴正正半轴上两点0,<a,<b,C(x,0)是X轴正半轴上任意一点,求当C在何位置时角ACB最大

设直线AC和BC的倾斜角分别为α, β.
tanα = (a-0)/(0-x) = -a/x
tanβ = (b-0)/(0-x) = -b/x
显然角ACB一定是锐角 = α-β
f(x)= tan(α- β) = (tanα - tanβ)/(1 + tanα*tanβ) = (-a/x + b/x)/(1 + ab/x²)= (b-a)x/(x²+ab)
f'(x) = (b-a)[1/(x²+ ab) + x*(-1)*2x/(x²+ab)²]
= (b-a)(ab - x²)/(x²+ab)² = 0
x = √(ab)
C(√(ab), 0)时角ACB最大

 解：设ｃ的坐标为（c,0) tanaco=a/c tanbco=b/c tan（ａｃｂ）＝tab(aco-bco)=(a/c-b/c)/(1+ab/c^2)=(a-b)/(c+ab/c) <=(a-b)/[2√(ab)] 当c=ab/c即c=√(ab)时，角ａｃｂ最大，此时ｃ坐标为(√(ab),0)  

设直线AC和BC的倾斜角分别为α, β. tanα = (a-0)/(0-x) = -a/x tanβ = (b-0)/(0-x) = -b/x 显然角ACB一定是锐角 = α-β f(x)= tan(α- β) = (tanα - tanβ)/(1 + tanα*tanβ) = (-a/x + b/x)/(1 + ab/x²)= (b-a)x/(x²+ab) f'(x) = (b-a)[1/(x²+ ab) + x*(-1)*2x/(x²+ab)²] = (b-a)(ab - x²)/(x²+ab)² = 0 x = √(ab) C(√(ab), 0)时角ACB最大