Answer
The critical points and their nature:
1. $\left( {0,0} \right)$, a saddle point.
2. $\left( { - \frac{1}{4},\frac{1}{2}} \right)$, a local minimum.
3. $\left( {\frac{{13}}{{64}}, - \frac{{13}}{{32}}} \right)$, a local minimum.
Work Step by Step
We have $f\left( {x,y} \right) = 8{y^4} + {x^2} + xy - 3{y^2} - {y^3}$.
The partial derivatives are
${f_x} = 2x + y$, ${\ \ \ }$ ${f_y} = 32{y^3} + x - 6y - 3{y^2}$
${f_{xx}} = 2$, ${\ \ }$ ${f_{yy}} = 96{y^2} - 6 - 6y$, ${\ \ }$ ${f_{xy}} = 1$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
$2x + y = 0$, ${\ \ \ }$ $32{y^3} + x - 6y - 3{y^2} = 0$
From the first equation we obtain $x = - \frac{y}{2}$. Substituting it in the second equation gives
$32{y^3} - \frac{y}{2} - 6y - 3{y^2} = 0$
$64{y^3} - 13y - 6{y^2} = 0$
$y\left( {64{y^2} - 6y - 13} \right) = 0$
So, the solutions are $y=0$, $y = \frac{1}{2}$ and $y = - \frac{{13}}{{32}}$.
Since $x = - \frac{y}{2}$, so the critical points are $\left( {0,0} \right)$, $\left( { - \frac{1}{4},\frac{1}{2}} \right)$ and $\left( {\frac{{13}}{{64}}, - \frac{{13}}{{32}}} \right)$.
Next, we use the contour map in Figure 20 to determine the nature of the critical points:
1. Critical point $\left( {0,0} \right)$
If we walk in the directions $ + {\bf{i}}$ or $ - {\bf{i}}$, it will take us uphill. If we walk in the directions $ + {\bf{j}}$ or $ - {\bf{j}}$, it will take us downhill. Therefore, we conclude that it is a saddle point.
2. Critical point $\left( { - \frac{1}{4},\frac{1}{2}} \right)$
It is a local minimum, because no matter which direction we walk, it will take us uphill.
3. Critical point $\left( {\frac{{13}}{{64}}, - \frac{{13}}{{32}}} \right)$
It is a local minimum, because no matter which direction we walk, it will take us uphill.
