Answer
The global minimum is $f\left( { - 5,3} \right) = - 34$ and the global maximum is $f\left( {5,3} \right) = 16$.
Work Step by Step
We have $f\left( {x,y} \right) = 5x - 3y$ and the domain $y \ge x - 2$, $y \ge - x - 2$, $y \le 3$.
Step 1. Find the critical points on the domain and evaluate $f$ at these points
The partial derivatives are
${f_x} = 5$, ${\ \ \ \ }$ ${f_y} = - 3$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$.
Since ${f_x} = 5$ and ${f_y} = - 3$, they will never become zero. Hence, there is no critical point of $f$.
Step 2. Check the boundaries
We restrict the function $f$ along the edges and find the minimum and maximum values. The results are given in the following table:
$\begin{array}{*{20}{c}}
{}&{{\rm{Restriction{\ }of}}}\\
{{\rm{Edge}}}&{f\left( {x,y} \right){\rm{to{\ }Edge}}}\\
{Top: - 5 \le x \le 5,y = 3}&{h\left( x \right) = f\left( {x,3} \right) = 5x - 9}\\
{Left: - 5 \le x \le 0,y = - x - 2}&{m\left( x \right) = f\left( {x, - x - 2} \right) = 8x + 6}\\
{Right:0 \le x \le 5,y = x - 2}&{n\left( x \right) = f\left( {x,x - 2} \right) = 2x + 6}
\end{array}\begin{array}{*{20}{c}}
{{\rm{Minimum{\ }of}}}&{{\rm{Maximum{\ }of}}}\\
{f\left( {x,y} \right){\rm{on{\ }Edge}}}&{f\left( {x,y} \right){\rm{on{\ }Edge}}}\\
{ - 34}&{16}\\
{ - 34}&6\\
6&{16}
\end{array}$
Step 3. Compare the results
From this table we obtain the global minimum $f\left( { - 5,3} \right) = - 34$ and the global maximum $f\left( {5,3} \right) = 16$.
