Lagrange multpliers

  • Thread starter Pi Face
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  • #1
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Use lagrange multipliers to find max/min of
f(x,y)=x^2+6y
subject to
x^2-y^2=5

grad f =λgrad g
2x=2xλ, λ=1
6=-2yλ, λ=-3/y
1=-3/y, y=-3

x^2-(-3)^2=x^2-9=5
x^2=14
x=+/-√14
two points are √14, -3 and -√14, -3
plugging both points into f(x,y) gives me the same answer. now what?
 

Answers and Replies

  • #2
Ray Vickson
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Use lagrange multipliers to find max/min of
f(x,y)=x^2+6y
subject to
x^2-y^2=5

grad f =λgrad g
2x=2xλ, λ=1
6=-2yλ, λ=-3/y
1=-3/y, y=-3

x^2-(-3)^2=x^2-9=5
x^2=14
x=+/-√14
two points are √14, -3 and -√14, -3
plugging both points into f(x,y) gives me the same answer. now what?


There are two constrained minima but no constrained maxima. Why not? Well, we can find values of x and y that together go to infinity along the curve g = 0, and when we do that we get f --> +infinity.

RGV
 
  • #3
76
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thank you very much
 

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