Question 877011
THE SOLUTION:
The region limited by the constraints {{{system(red(y-x>=1),green(2<=x<=5),blue(y-x<=3))}}} (it needs to be {{{blue(y-x<=3)}}} to have a solution)
is the parallelogram below, with vertices at (2,3), (2,5), (5,6), and (5,8).
{{{drawing(450,450,-4,8,-1,9,
grid(1),red(line(-6,-5,14,15)),
green(line(2,-6,2,14)),green(line(5,-6,5,14)),
blue(line(-6,-3,14,17))
)}}} The maximum is at (5,8), where {{{F=9*5+40*8=45+320=highlight(365)}}} .
The minimum is at (2,3), where {{{F=9*2+40*3=18+120=highlight(138)}}} .
You would know how much you have to "show your work,"
I will just over-explain how to get to the solution, because I do not know how much explanation you want/need.
 
HOW TO GET TO THE SOLUTION:
I like to make a graph/sketch, because it helps me understand the problem, and avoid mistakes.
Graphing the "feasible region" limited by those constraints is nice and probably expected.
However, this region was so easy to imagine and calculate that graphing was not absolutely needed.
The constraints set the limits/borders of the feasible region as portions of the lines {{{system(red(y-x=1),green(2=x),green(x=5),blue(y-x=3))}}} .
We should realize that {{{green(2=x)}}} , and {{{green(x=5)}}} are vertical (and parallel) lines,
and that {{{red(y-x=1)}}} , and {{{blue(y-x=3)}}} are slanted (and parallel) lines.
The feasible region has four parallel sides; it is a parallelogram.
By plotting all four of those lines on the same graph, I find the corners graphically.
Otherwise, I would have to solve fours systems of two equations,
like {{{system(red(y-x=1),green(2=x))}}} ,
to find the corners where those lines intersect.
How to graph the vertical lines {{{green(2=x)}}} and {{{green(x=5)}}} is obvious.
The region limited by {{{green(2<=x<=5)}}} is obviously the space between those vertical lines, where {{{x}}} is at least {{{2}}}, but less than {{{5}}} .
Graphing {{{red(y-x=1)}}} and {{{blue(y-x=3)}}} is a tiny bit more difficult, but still easy.
To graph {{{red(y-x=1)}}} and {{{blue(y-x=3)}}} , we can transform the equations into the equivalent slope-intercept form,
{{{red(y-x=1)}}} ---> {{{red(y=x+1)}}} 
{{{blue(y-x=3)}}} ---> {{{blue(y=x+3)}}} .
{{{blue(y=x+3)}}} <-->{{{blue(y=1*x+3)}}} tells us that
the blue line has a y-intercept of {{{3}}} ,
and a slope of {{{1}}} , meaning that as x increases by 1, y increases by 1,
so the blue line crosses the y-axis at (0,3), and from there goes to (1,4), (2,5), (3,8), and so on.
Otherwise, we could find the x- and y-intercepts by respectively setting
{{{y=0}}} (which is true for all points on the x-axis) and
{{{x=0}}} (which is true for all points on the y-axis).
For {{{y=0}}} , {{{blue(y-x=3)}}} becomes {{{blue(0-x=3)}}}-->{{{blue(-x=3)}}}-->{{{blue(x=-3)}}} ,
so we know that the x-intercept for {{{blue(y-x=3)}}} is (-3,0),
and for {{{x=0}}} , {{{blue(y-x=3)}}} becomes {{{blue(y-0=3)}}}-->{{{blue(y=3)}}} ,
so we know that the y-intercept for {{{blue(y-x=3)}}} is (0,3).
With those intercepts we can connect (-3,0) and (0,3) with a straight line to get the graph of {{{blue(y-x=3)}}} .
The constraints {{{system(red(y-x>=1),blue(y-x<=3))}}} <---> {{{system(red(y>=x+1),blue(y<=x+3))}}} determine the region between the red and blue lines.
We know that because
we notice that (0,0) with {{{x=0}}} and {{{y=0}}} is a solution to {{{system(red(y-x>=1),blue(y-x<=3))}}} ,
or because we realize that {{{system(red(y>=x+1),blue(y<=x+3))}}} is the space above the red line {{{red(y=x+1)}}} , but below the blue line {{{blue(y=x+3)}}} .
 
Once we have determined that the feasible region is the parallelogram bordered by those lines,
we determine the value of {{{F = 9x + 40y}}} at the corners of the feasible region.
In any problem of this kind, a linear function of x and y, like {{{F = 9x + 40y}}} ,
will be maximum at one corner or at two corners and a border line connecting those two corners.
The same goes for the minimum.
In this case, it was obvious that
the maximum values for x and y in the feasible region, at (5,8), would give the maximum value to {{{F = 9x + 40y}}} ,
and that the minimum values for x and y, at (2,3), would give the minimum value to {{{F = 9x + 40y}}} .
In general, you would have to show the calculations for all corners:
at (2,3), {{{F=9*2+40*3=18+120=138}}} ,
at (2,5), {{{F=9*2+40*5=18+200=218}}} ,
at (5,6), {{{F=9*5+40*6=45+240=285}}} ,
at (5,8), {{{F=9*5+40*8=45+320=365}}} ,
and then you would compare the values to find that
the greatest (largest) value, {{{F=9*5+40*8=45+320=highlight(365)}}} , is the maximum,
and the smallest value, {{{F=9*2+40*3=18+120=highlight(138)}}} , is the minimum.