<s>[INST] maximum and minimum of $x^2+y^4$ for real $x,y$ If $y^2(y^2-6)+x^2-8x+24=0$ then maximum and minimum value of $x^2+y^4$ is 
what i try
$y^4-6y^2+9+x^2-8x+16=1$
$(x-4)^2+(y^2-3)^2=1\cdots (1)$
How i find maximum and minimum of $x^2+y^4$ from $(1)$ relation
help me to solve it please 
 [/INST] By C-S $$0=y^2(y^2-6)+x^2-8x+24=$$
$$=x^2+y^4-8x-6y^2+24\geq$$
$$\geq x^2+y^4-\sqrt{(8^2+6^2)(x^2+y^4)}+24=$$
$$=x^2+y^4-10\sqrt{x^2+y^4}+25-1=\left(\sqrt{x^2+y^4}-5\right)^2-1,$$
which gives
$$4\leq\sqrt{x^2+y^4}\leq6$$ and
$$16\leq x^2+y^4\leq36.$$
The equality occurs for $(|x|,y^2)||(8,6),$ which says that $36$ is a maximal value and $16$ is a minimal value.
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