Answer
The global minimum of $f$ is $f\left( { - 2,4} \right) = - 18$ and the global maximum is $f\left( {2,4} \right) = 10$.
Work Step by Step
We have $f\left( {x,y} \right) = 2xy - x - y$ and the domain $\left\{ {y \le 4,y \ge {x^2}} \right\}$.
Step 1. Find the critical points on the domain and evaluate $f$ at these points
The partial derivatives are
${f_x} = 2y - 1$, ${\ \ \ }$ ${f_y} = 2x - 1$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$.
$2y-1=0$, ${\ \ \ }$ $2x-1=0$
So, there is a critical point at $\left( {\frac{1}{2},\frac{1}{2}} \right)$.
The extreme value corresponding to this critical point is $f\left( {\frac{1}{2},\frac{1}{2}} \right) = - \frac{1}{2}$.
Step 2. Check the boundaries
1. On the top edge of the domain, $y=4$ and $ - 2 \le x \le 2$. Write $g\left( x \right) = f\left( {x,4} \right) = 7x - 4$.
Thus, at the top edge the minimum value of $f$ is $f\left( { - 2,4} \right) = - 18$ and the maximum value of $f$ is $f\left( {2,4} \right) = 10$.
2. On the left edge of the domain, $y = {x^2}$ and $ - 2 \le x \le 0$. Write $h\left( x \right) = f\left( {x,{x^2}} \right) = 2{x^3} - {x^2} - x$.
Find the extreme values of $f$ on the left edge.
The critical points of $h$ can be found by solving the equation $h'\left( x \right) = 0$.
$h'\left( x \right) = 6{x^2} - 2x - 1 = 0$, ${\ \ \ }$ $h{\rm{''}}\left( x \right) = 12x - 2$
$x = \frac{{2 \pm \sqrt {{{\left( { - 2} \right)}^2} - 4\cdot6\cdot\left( { - 1} \right)} }}{{2\cdot6}} = \frac{{2 \pm \sqrt {28} }}{{12}} = \frac{{1 \pm \sqrt 7 }}{6}$
For $ - 2 \le x \le 0$, the critical point of $h$ is $x = \frac{{1 - \sqrt 7 }}{6} \simeq - 0.274$. Since $h{\rm{''}}\left( {\frac{{1 - \sqrt 7 }}{6}} \right) < 0$, $h\left( x \right)$ has a maximum at $x = \frac{{1 - \sqrt 7 }}{6}$. So, the maximum of $f$ along the left edge is $h\left( {\frac{{1 - \sqrt 7 }}{6}} \right) = f\left( {\frac{{1 - \sqrt 7 }}{6},{{\left( {\frac{{1 - \sqrt 7 }}{6}} \right)}^2}} \right) \simeq 0.158$.
Since $h\left( x \right)$ has a maximum, the curve is concave down. Thus, for $ - 2 \le x \le 0$ the minimum of $f$ occurs at $x=-2$. So, the minimum of $f$ along the left edge is $h\left( { - 2} \right) = f\left( { - 2,4} \right) = - 18$.
3. On the right edge of the domain, $y = {x^2}$ and $0 \le x \le 2$. Write $m\left( x \right) = f\left( {x,{x^2}} \right) = 2{x^3} - {x^2} - x$.
Find the extreme values of $f$ on the right edge.
The critical points of $m$ can be found by solving the equation $m'\left( x \right) = 0$.
$m'\left( x \right) = 6{x^2} - 2x - 1 = 0$, ${\ \ \ }$ $m{\rm{''}}\left( x \right) = 12x - 2$
$x = \frac{{2 \pm \sqrt {{{\left( { - 2} \right)}^2} - 4\cdot6\cdot\left( { - 1} \right)} }}{{2\cdot6}} = \frac{{2 \pm \sqrt {28} }}{{12}} = \frac{{1 \pm \sqrt 7 }}{6}$
For $0 \le x \le 2$, the critical point of $m$ is $x = \frac{{1 + \sqrt 7 }}{6} \simeq 0.608$. Since $m{\rm{''}}\left( {\frac{{1 + \sqrt 7 }}{6}} \right) > 0$, $m\left( x \right)$ has a minimum at $x = \frac{{1 + \sqrt 7 }}{6}$. So, the minimum of $f$ along the right edge is $m\left( {\frac{{1 + \sqrt 7 }}{6}} \right) = f\left( {\frac{{1 + \sqrt 7 }}{6},{{\left( {\frac{{1 + \sqrt 7 }}{6}} \right)}^2}} \right) \simeq - 0.528$.
Since $m\left( x \right)$ has a minimum, the curve is concave up. Thus, for $0 \le x \le 2$ the maximum of $f$ occurs at $x=2$. So, the maximum of $f$ along the right edge is $m\left( 2 \right) = f\left( {2,4} \right) = 10$.
The extrema on the edges are summarized in the following table:
$\begin{array}{*{20}{c}}
{}&{{\rm{Restriction{\ }of}}}\\
{{\bf{Edge}}}&{f\left( {x,y} \right){\rm{to{\ }Edge}}}\\
{Top:y = 4, - 2 \le x \le 2}&{g\left( x \right) = f\left( {x,4} \right) = 7{\rm{x}} - 4}\\
{Left:y = {x^2}, - 2 \le x \le 0}&{h\left( x \right) = f\left( {x,{x^2}} \right) = 2{x^3} - {x^2} - x}\\
{Right:y = {x^2},0 \le x \le 2}&{m\left( x \right) = f\left( {x,{x^2}} \right) = 2{x^3} - {x^2} - x}
\end{array}\begin{array}{*{20}{c}}
{{\rm{min{\ }of}}}&{{\rm{max{\ }of}}}\\
{{\rm{f{\ }on{\ }Edge}}}&{{\rm{f{\ }on{\ }Edge}}}\\
{ - 18}&{10}\\
{ - 18}&{0.158}\\
{ - 0.528}&{10}
\end{array}$
Step 3. Compare the results
Comparing our results in Step 1 and Step 2, we conclude that the global minimum of $f$ is $f\left( { - 2,4} \right) = - 18$ and the global maximum is $f\left( {2,4} \right) = 10$.
