以实数x,y满足x²+y²-4x+1=0 （1）求x÷y的最大值和最小值
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x^2+y^2-4x+1=0
2x+ 2y.dy/dx - 4 =0
dy/dx = (2-x)/y
S = x/y
dS/dx = [y - x(dy/dx) ] /y^2
dS/dx =0
y - x(dy/dx) =0
y- x(2-x)/y =0
y^2 - 2x + x^2 =0
(-x^2+4x-1) - 2x + x^2 =0
2x-1=0
x=1/2
x=1/2
x^2+y^2-4x+1=0
(1/4)+y^2-2+1=0
y^2 = 3/4
y = √3/2 or -√3/2
max x/y = (1/2)/(√3/2) = √3/3
2x+ 2y.dy/dx - 4 =0
dy/dx = (2-x)/y
S = x/y
dS/dx = [y - x(dy/dx) ] /y^2
dS/dx =0
y - x(dy/dx) =0
y- x(2-x)/y =0
y^2 - 2x + x^2 =0
(-x^2+4x-1) - 2x + x^2 =0
2x-1=0
x=1/2
x=1/2
x^2+y^2-4x+1=0
(1/4)+y^2-2+1=0
y^2 = 3/4
y = √3/2 or -√3/2
max x/y = (1/2)/(√3/2) = √3/3
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