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Expert-verified Found in: Page 985 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # In Exercises, by considering the function $f\left(x,y\right)={x}^{2}y$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 $\left(0,0\right)$is not an extremum of $f\left(x,y\right)={x}^{2}y$subject to the constraint $x+y=0.$

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

See the step by step solution

## Step 1. Given information.

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

Given constraint is $x+y=0.$

## Step 2. Explanation.

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

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

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 $f\left({x}_{0},{y}_{0}\right)<0.$

The function is negative as well as positive for the neighbor points of the point $\left(0,0\right)$so the point $\left(0,0\right)$can not be an extremum of the function $f\left(x,y\right)={x}^{2}y$ ### Want to see more solutions like these? 