I am finding the extreme points of the function z = x 3 </msup> +

realburitv4

realburitv4

Answered question

2022-05-28

I am finding the extreme points of the function z = x 3 + y 3 ( x + y ) 2 . This function has two critical points (0;0) and (4/3;4/3). The second point is an extreme point. For the first point, I have no conclusion. I claimed it is not an extreme point. Clear, z < 0 when x , y < 0. I am looking for ( x ; y ) such that z > 0. Am I correct?

Answer & Explanation

Porter Cohen

Porter Cohen

Beginner2022-05-29Added 7 answers

The quadratic form in the Maclaurin expansion, Q ( x , y ) = ( x + y ) 2 , is negative semidefinite, so the second derivative test is inconclusive. Then it's usually a good idea to see what happens at the points where Q = 0, since that's where the higher-order term may have a chance of making their voice heard.

So have a look at what happens along the line y = x:
f ( t , t ) =

EDIT: Have a look at what happens along a curve which is tangent to the line y = x at the origin:
f ( t , t + t 2 ) =

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