Lagrange multipliers

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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

Oct 29th, 2015

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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

Please let me know if you need any clarification. I'm always happy to answer your questions.
Oct 29th, 2015

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