Answer
(a) Setting ${f_x}\left( {x,y} \right) = 0$ and ${f_y}\left( {x,y} \right) = 0$, we obtain the critical points:
$P = \left( {0,0} \right)$, $\left( {2\sqrt 2 ,\sqrt 2 } \right)$, $\left( { - 2\sqrt 2 , - \sqrt 2 } \right)$.
(b) The local minima are at $\left( {2\sqrt 2 ,\sqrt 2 } \right)$ and $\left( { - 2\sqrt 2 , - \sqrt 2 } \right)$.
The saddle point is at $\left( {0,0} \right)$.
The absolute minimum value is $-4$.
Work Step by Step
(a) We have $f\left( {x,y} \right) = {x^2} + {y^4} - 4xy$.
The partial derivatives are
${f_x} = 2x - 4y$, ${\ \ \ }$ ${f_y} = 4{y^3} - 4x$
We find the critical points by first setting ${f_x}\left( {x,y} \right) = 0$.
$2x - 4y = 0$, ${\ \ \ }$ $x = 2y$
Write $x=a$ and $y=b$, then $a=2b$. So, the solution is $\left( {x,y} \right) = \left( {2b,b} \right)$.
Setting ${f_y}\left( {x,y} \right) = 0$ we get $4{y^3} - 4x = 0$.
Substituting $\left( {x,y} \right) = \left( {2b,b} \right)$ in this equation gives
$4{b^3} - 8b = 0$
$b\left( {{b^2} - 2} \right) = 0$
The solutions are $b=0$ or $b = \pm \sqrt 2 $.
Hence, the critical points are $P = \left( {0,0} \right)$, $\left( {2\sqrt 2 ,\sqrt 2 } \right)$, $\left( { - 2\sqrt 2 , - \sqrt 2 } \right)$.
(b) From Figure 18, we see that the local minima are at $\left( {2\sqrt 2 ,\sqrt 2 } \right)$ and $\left( { - 2\sqrt 2 , - \sqrt 2 } \right)$.
The saddle point is at $\left( {0,0} \right)$.
The absolute minimum value is obtained by substituting either $\left( {2\sqrt 2 ,\sqrt 2 } \right)$ or $\left( { - 2\sqrt 2 , - \sqrt 2 } \right)$ in $f\left( {x,y} \right)$:
$f\left( {2\sqrt 2 ,\sqrt 2 } \right) = f\left( { - 2\sqrt 2 , - \sqrt 2 } \right) = - 4$
So, the absolute minimum value is $-4$.
