Answer
1. $f\left( {x,y} \right) = {x^2} + 2{y^2} - 4y + 6x$.
Critical point: $\left( {x,y} \right) = \left( { - 3,1} \right)$
$f\left( { - 3,1} \right)$ is a local minimum.
$f\left( {x,y} \right)$ matches graph (B) in Figure 19.
2. $g\left( {x,y} \right) = {x^2} - 12xy + y$.
Critical point: $\left( {x,y} \right) = \left( {\frac{1}{{12}},\frac{1}{{72}}} \right)$.
$\left( {\frac{1}{{12}},\frac{1}{{72}}} \right)$ is a saddle point.
$g\left( {x,y} \right)$ matches graph (A) in Figure 19.
In summary:
$\begin{array}{*{20}{c}}
{{\bf{Function}}}&{{\bf{Critical{\ }point}}}&{{\bf{Type}}}&{{\bf{Match}}}\\
{f\left( {x,y} \right) = {x^2} + 2{y^2} - 4y + 6x}&{\left( { - 3,1} \right)}&{{\rm{local{\ }minimum}}}&{{\rm{graph }}\left( {\rm{B}} \right)}\\
{g\left( {x,y} \right) = {x^2} - 12xy + y}&{\left( {\frac{1}{{12}},\frac{1}{{72}}} \right)}&{{\rm{saddle{\ }point}}}&{{\rm{graph }}\left( {\rm{A}} \right)}
\end{array}$
Work Step by Step
1. We have $f\left( {x,y} \right) = {x^2} + 2{y^2} - 4y + 6x$.
The partial derivatives are
${f_x} = 2x + 6$, ${\ \ \ }$ ${f_y} = 4y - 4$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
$2x + 6 = 0$, ${\ \ \ }$ $x=-3$
$4y-4=0$, ${\ \ \ }$ $y=1$
So, there is only one critical point at $\left( {x,y} \right) = \left( { - 3,1} \right)$.
The Second Derivative Test:
We evaluate the partial derivatives
${f_{xx}} = 2$, ${\ \ }$ ${f_{yy}} = 4$, ${\ \ }$ ${f_{xy}} = 0$
The discriminant at $\left( {x,y} \right) = \left( { - 3,1} \right)$ is
$D = {f_{xx}}{f_{yy}} - {f_{xy}}^2 = 8$
Since $D > 0$ and ${f_{xx}} > 0$, by the Second Derivative Test, $f\left( { - 3,1} \right)$ is a local minimum.
Thus, $f\left( {x,y} \right)$ matches graph (B) in Figure 19.
2. We have $g\left( {x,y} \right) = {x^2} - 12xy + y$.
The partial derivatives are
${g_x} = 2x - 12y$, ${\ \ \ }$ ${g_y} = - 12x + 1$
To find the critical points we solve the equations ${g_x} = 0$ and ${g_y} = 0$:
$2x - 12y = 0$, ${\ \ \ }$ $ - 12x + 1 = 0$
From the second equation we obtain $x = \frac{1}{{12}}$. Substituting it in the first gives $y = \frac{1}{{72}}$.
So, there is only one critical point at $\left( {x,y} \right) = \left( {\frac{1}{{12}},\frac{1}{{72}}} \right)$.
The Second Derivative Test:
We evaluate the partial derivatives
${g_{xx}} = 2$, ${\ \ }$ ${g_{yy}} = 0$, ${\ \ }$ ${g_{xy}} = - 12$
The discriminant at $\left( {x,y} \right) = \left( {\frac{1}{{12}},\frac{1}{{72}}} \right)$ is
$D = {g_{xx}}{g_{yy}} - {g_{xy}}^2 = - 144$
Since $D < 0$, by the Second Derivative Test, $\left( {\frac{1}{{12}},\frac{1}{{72}}} \right)$ is a saddle point.
Thus, $g\left( {x,y} \right)$ matches graph (A) in Figure 19.
