use the method of Lagrange multipliers to find the extrema of the function f subject to the given conatraint.
f(x,y)= xy ; x^2+y^2=1
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F(x,y） = xy-L(x^2+y^2-1)
dG/dx=0 ---> y - 2xL =0 y = 2xL
dG/dy =0 ---> x - 2yL = 0
y/x = x/y
y^2 = x^2
x = +- sqrt(2)/2
y = +-sqrt(2)/2
maximum = 1/2
x = y = sqrt(2)/2 or x=y=-sqrt(2)/2
minimum = -1/2
x = -y = sqrt(2)/2 or x = -y = -sqrt(2)/2
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