Find and sort all of the extremum of $f(x,y) = |3x| + x^4y^2$

not quite sure what to do with the absolute value of $|3x|$ , should i separate it to two options? , one for $3x>0$ and the other option for $3x<0$ ?

  • from $f(x,y) = |3x| + x^4y^2= 3|x| + |x|^4|y|^2$ I think you can discuss with $x\geq0$ and $y\geq0$ in $f(x,y) = 3x + x^4y^2$. – Nosrati Sep 14 at 13:19
  • how do i differentiate it? $f_x = 3+4x^3y^2 $ ? am i supposed to just ignore the absolute value? – Maor Rocky Sep 14 at 13:24
  • Its minimum value is $0$, all along the $y$-axis. – Empy2 Sep 14 at 13:35
up vote 1 down vote accepted

from $f(x,y) = |3x| + x^4y^2= 3|x| + |x|^4|y|^2$ I think you can discuss with $x\geq0$ and $y\geq0$ in $f(x,y) = 3x + x^4y^2$.

$f_x=3+4x^3y^2=0$

$f_y=2x^4y=0$

with the second $x=0$ or $y=0$ which both are impossible as conflict with the first, remains $$f(x,y) = |3x| + x^4y^2= 3|x| + |x|^4|y|^2\geq0=f(0,0)$$ specifies the only minimum point in the origin.

Hint: it is clear that $$3|x|+x^4y^2\geq 0$$ so the Minimum is given by $$(0,0)$$

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