Question 877011: Find the maximum and minimum value of F = 9x + 40y subject to the constraints
a) y-x≥1
b) 2≤x≤5
c) y-x<3
Can you please help me out ? Thanks so much in advance:)
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! THE SOLUTION:
The region limited by the constraints  (it needs to be  to have a solution)
is the parallelogram below, with vertices at (2,3), (2,5), (5,6), and (5,8).
 The maximum is at (5,8), where  .
The minimum is at (2,3), where  .
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  .
We should realize that  , and  are vertical (and parallel) lines,
and that  , and  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  ,
to find the corners where those lines intersect.
How to graph the vertical lines and is obvious.
The region limited by  is obviously the space between those vertical lines, where  is at least  , but less than  .
Graphing and is a tiny bit more difficult, but still easy.
To graph and , we can transform the equations into the equivalent slope-intercept form,
---> 
--->  .
<--> tells us that
the blue line has a y-intercept of  ,
and a slope of  , 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
 (which is true for all points on the x-axis) and
 (which is true for all points on the y-axis).
For , becomes  --> --> ,
so we know that the x-intercept for is (-3,0),
and for , becomes  --> ,
so we know that the y-intercept for is (0,3).
With those intercepts we can connect (-3,0) and (0,3) with a straight line to get the graph of .
The constraints  <--->  determine the region between the red and blue lines.
We know that because
we notice that (0,0) with and is a solution to ,
or because we realize that is the space above the red line , but below the blue line .
Once we have determined that the feasible region is the parallelogram bordered by those lines,
we determine the value of  at the corners of the feasible region.
In any problem of this kind, a linear function of x and y, like ,
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 ,
and that the minimum values for x and y, at (2,3), would give the minimum value to .
In general, you would have to show the calculations for all corners:
at (2,3),  ,
at (2,5),  ,
at (5,6),  ,
at (5,8),  ,
and then you would compare the values to find that
the greatest (largest) value, , is the maximum,
and the smallest value, , is the minimum.
|
|
|
