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Chapter 12: Q. 15 (page 985)

In Exercises, by considering the function f(x,y)=x2ysubject 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)=x2ysubject to the constraintx+y=0.

Short Answer

Expert verified

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 functionf(x,y)=x2y.

Step by step solution

01

Step 1. Given information.      

The given function isf(x,y)=x2y.

Given constraint isx+y=0.

02

Step 2. Explanation.

Consider point x0,y0is the neighbor point of (0,0)where y0>0.

Then the function is positive at (x0,y0)so that f(x0,y0)>0.

Consider point x1,y1is the neighbor point of 0,0where y0<0.

Then the function is negative at x1,y1so that f(x0,y0)<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)=x2y

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