- #1

- 76

- 0

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?

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- Thread starter Pi Face
- Start date

- #1

- 76

- 0

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?

- #2

Ray Vickson

Science Advisor

Homework Helper

Dearly Missed

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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

- 0

thank you very much

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