In Exercises, by considering the function $f (x, y) = x^{2} y$ subject to the constraint $x + y = 0,$ you will explore a situation in which the method of Lagrange multipliers does not provide an extremum of a function.
Explain why $(0, 0)$ is not an extremum of $f (x, y) = x^{2} y$ subject to the constraint $x + y = 0 .$

Question

In Exercises, by considering the function $f (x, y) = x^{2} y$ subject to the constraint $x + y = 0,$ you will explore a situation in which the method of Lagrange multipliers does not provide an extremum of a function.
Explain why $(0, 0)$ is not an extremum of $f (x, y) = x^{2} y$ subject to the constraint $x + y = 0 .$

Accepted Answer

The function is negative as well as positive for the neighbor points of the point $(0, 0)$ so the point $(0, 0)$ can not be an extremum of the function $f (x, y) = x^{2} y .$

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In Exercises, by considering the function f(x,y)=x2y subject to the constraint x+y=0, you will explore a situation in which the method of Lagrange multipliers does not provide an extremum of a function.Explain why (0,0) is not an extremum of f(x,y)=x2y subject to the constraint x+y=0.

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