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Q. 15

Expert-verifiedFound in: Page 985

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

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 $x+y=0.$

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 $f(x,y)={x}^{2}y.$

The given function is $f(x,y)={x}^{2}y.$

Given constraint is $x+y=0.$

Consider point $\left({x}_{0},{y}_{0}\right)$is the neighbor point of $(0,0)$where ${y}_{0}>0.$

Then the function is positive at $({x}_{0},{y}_{0})$so that $f({x}_{0},{y}_{0})>0.$

Consider point $\left({x}_{1},{y}_{1}\right)$is the neighbor point of $\left(0,0\right)$where ${y}_{0}<0.$

Then the function is negative at $\left({x}_{1},{y}_{1}\right)$so that $f({x}_{0},{y}_{0})<0.$

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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