Question 1026330
{{{x<=2}}}
{{{3x-y>=-2}}}<--->{{{3x+2>=y}}}
{{{y>=x-2}}}
 
We can graph the boundary lines,
{{{red(x=2)}}} , {{{blue(y=3x+2)}}} , and {{{green(y=x-2)}}} .
{{{drawing(200,300,-5,5,-5,10,
grid(1),red(line(2,-5,2,10)),
blue(line(-3,-7,3,11)),
green(line(-3,-5,5,3))
)}}}
We can find their intersections points (the vertices of the feasible region)
by solving each 2-line system of equations:
{{{system(x=2,y=x-2)}}}--->{{{highlight(system(x=2,y=0))}}}
{{{system(x=2,y=3x+2)}}}--->{{{highlight(system(x=2,y=8))}}}
{{{system(y=3x+2,y=x-2)}}}--->{{{highlight(system(x=-2,y=-4))}}} .
The function will be minimum at one of those vertices or all along the boundary line that joins two of them.
Same thing goes for the maximum.
So, we calculate the value for {{{f(x, y) = 2x-3y}}} ate each one of those vertices:
{{{f(2, 0) = 2*2-3*0=4}}}
{{{f(2, 8) = 2*2-3*8=4-24=-20}}}
{{{f(-2, -4) = 2(-2)-3(-4)=-4+12=8}}} .
The minimum is {{{highlight(-20)}}} .
The maximum is {{{highlight(8)}}} .