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Question: $x^2 + 4xy + 4y^2 + x^3 +2x^2y+y^4 $ critical point $(0,0)$, $F_{xx}*F_{yy} - (F_{xy})^2$ = $0$ now test failed to conclude any thing. But solution says it has minimum at the origin. Please guide me how to proceed. And lastly, I am unable to find its other critical point. Please help.

VIVID
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Abhishek Verma
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    Are you able to prove that the function is always positive for all x and all y? – Ishraaq Parvez Jan 01 '21 at 09:43
  • after simplyfying $(x+2y)^2 +x^2 (x+2y) +y^4$, now in close NBD of (0,0) ,1>x^2 then (x+2y) >$ x^2( x+2y) $ then definitely $(x+2y)^2 > x^2(x+2y)$ hence the given expression is greater then zero for all (x,y). Please see if it's correct – Abhishek Verma Jan 01 '21 at 11:31
  • Try to prove that the function f(x,y)=x^3+2x^2y has a minima at the origin. Since we know that the rest of the given funtion is always positive for all x and y. – Ishraaq Parvez Jan 02 '21 at 03:22

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Note that

$$ p(x,y) = x^2 + 4 x y + 4 y^2 + x^3 + 2 x^2 y + x^4=\left(\begin{array}{c}y\\ x\\ x^2\end{array}\right)^{\dagger}\left(\begin{array}{ccc}4& 0 & 0\\ 4 & 1 & 0\\ 2 & 1 & 1\end{array}\right)\left(\begin{array}{c}y\\ x\\ x^2\end{array}\right) $$

Note also that matrix $M$

$$ M = \left(\begin{array}{ccc}4& 0 & 0\\ 4 & 1 & 0\\ 2 & 1 & 1\end{array}\right) $$

has eigenvalues $\{4,1,1\}$ then it is positive definite hence $p(x,y)$ has a minimum at the origin.

Cesareo
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