Critical points (x-1)^4 (x-y)^4

[TRoad]

Homework Statement

Find critical points. $f(x,y) = (x-1)^4 + (x-y)^4$

The Attempt at a Solution

$\frac{\partial f}{\partial x} = 4(x-1)^3 + 4(x-y)^3$

$\frac{\partial f}{\partial y} = - 4(x-y)^3$


$\frac{\partial f}{\partial x} = 0$,

$\frac{\partial f}{\partial y} = 0$,

$- 4(x-y)^3 = 0$, x=y

$4(y-1)^3 + 4(y-y)^3 = 0$,


We have a critical point at: y=1,x=1

So as the both part of the function ( $(x-1)^4$ and $(x-y)^4$ ) will always be grater or equeal 0, and the critical point gives us 0, then I guess it's a min, no?

So if I'm not wrong, then why is inconclusive the second derivative text AC-B^2=0

A=Fxx, C=Fyy, B=Fxy

Thanks!

 

[Ray Vickson]

Homework Statement

Find critical points. $f(x,y) = (x-1)^4 + (x-y)^4$

The Attempt at a Solution

$\frac{\partial f}{\partial x} = 4(x-1)^3 + 4(x-y)^3$

$\frac{\partial f}{\partial y} = - 4(x-y)^3$


$\frac{\partial f}{\partial x} = 0$,

$\frac{\partial f}{\partial y} = 0$,

$- 4(x-y)^3 = 0$, x=y

$4(y-1)^3 + 4(y-y)^3 = 0$,


We have a critical point at: y=1,x=1

So as the both part of the function ( $(x-1)^4$ and $(x-y)^4$ ) will always be grater or equeal 0, and the critical point gives us 0, then I guess it's a min, no?

So if I'm not wrong, then why is inconclusive the second derivative text AC-B^2=0

A=Fxx, C=Fyy, B=Fxy

Thanks!

The second derivatives are all zero at the stationary point, so second-order conditions give no information. However, the function f(x,y) is strictly convex, so any stationary point must be a global min.

BTW: having second-order things = 0 does not guarantee minimality: the *sufficient* second-order conditions need a strict inequality ">0". Another way to see this is to note that g(x,y) = -f(x,y) also has all second derivatives = 0, but for it the stationary point is a maximum!

 

[TRoad]

Understood! Thanks :)