Question 1207825: use graphical methods and the corner point theorem to maximize z=x+5y subject to 3x+2y<12
2x+y<7
x>0
y<0
Found 4 solutions by ikleyn, Edwin McCravy, greenestamps, mccravyedwin:
Answer by ikleyn(52833)   (Show Source):
You can put this solution on YOUR website! .
use graphical methods and the corner point theorem to maximize z=x+5y subject to 3x+2y<12
2x+y<7
x>0
y<0
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With the inequality signs written/presented in your post, function z = x + 5y has no maximum
in the feasibility domain.
So, the problem, as formulated in the post, does not have a solution, at all.
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The corner points theorem is inapplicable in this case, since the corner points,
defined by inequalities in your post, do not belong to the feasibility domain.
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In order for the problem be meaningful and would have a solution, the problem should be RE-formulated
and the inequality signs should be changed.
So, guided by these my notices, double check your post, comparing it with the source.
If the problem, as written in your post, coincide with the source, then the problem is a TRAP and has no solution.
In case, if you detect errors in your post, fix them and re-post the fixed version to the forum.
If you do it, then please do not post the problem to me in person - re-post as you usually post to this forum,
through the central gate.
As the lesson to learn for the future, memorize that Math does not tolerate inaccuracy.
Answer by Edwin McCravy(20060)  (Show Source):
You can put this solution on YOUR website!
What she means is that you should have ≤, not <, and ≥, not >.
Otherwise your inequalities will not include the boundary lines,
and will have no maximum (or minimum) solutions.
Maximize z=x+5y
subject to
3x+2y≤12
2x+y≤7
x≥0, y≥0
Here is the first quadrant graphs of the lines whose equations
are like the constraint inequalities with equal signs instead of
the inequality symbols, 3x+2y≤12 and 2x+y≤7 . Use the origin
(x,y) = (0,0) as a test point to find which side of each line the
shading will appear on.
Testing 3x+2y≤12: 3(0)+2(0)≤12 is true so we shade the side of that
line which the origin is on, which is underneath the line:
Testing 2x+y≤7: 2(0)+(0)≤7 is true so we shade the side of that
line which the origin is on, which is also underneath the line.
so we shade under BOTH lines.
The feasible region is the shaded area.
We find all the corner points. by finding the x and y intercepts of both
lines and their point of intersection.
The y-intercept of
3x+2y=12
is found by substituting x=0 and solving for y,
we get y-intercept for the red line is (0,6)
The x-intercept of
2x+y=7
is found by substituting y=0 and solving for x,
we get x-intercept for the blue line is (3.5,0)
We get the corner point which is the intersection of
the two lines:
Solve that by substitution or elimination and get (2,3)
Now we find the value of the objective function
z = x+5y at each of the 4 corner points:
At corner point (x,y) = (0,0), z = x+5y = 0+5(0) = 0.
At corner point (x,y) = (0,6), z = x+5y = 0+5(6) = 30.
At corner point (x,y) = (2,3), z = x+5y = 2+5(3) = 17.
At corner point (x,y) = (3.5,0), z = x+5y = 3.5+5(0) = 3.5.
So z has a maximum value of 30 when x=0 and y=6.
Edwin
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
Tutor @ikleyn provides a lengthy response stating that the problem as posed has no solution, without saying what the problem is.
Tutor @Edwin tells what the problem is with the problem as posted in his first sentence. A bit more helpful to the student....
His response shows a typical solution using the method found in virtually all references: determine the corners of the feasibility region based on the given constraints and evaluate the objective function at each corner to find the minimum and maximum values of the objective function.
In fact, it is not necessary to evaluate the objective function at each corner of the feasibility region.
You can determine the corners of the feasibility region where the minimum and maximum values of the objective function occur by comparing the slope of the objective function with the slopes of the constraint boundary lines.
The objective function is z=x+5y; in slope-intercept form it is y=(-1/5)x plus some constant, so the slope of the objective function is -1/5.
The maximum and minimum values of the objective function will occur where lines with slopes of -1/5 just touch the feasibility region.
The slopes of the two constraint boundary lines are -2 and -3/2.
Using the diagram Edwin shows in his post, it is clear that lines with slopes of -1/5 will just touch the feasibility region at (0,6) and (0,0), so those are the corners of the feasibility region where the maximum and minimum values of the objective function will occur.
ANSWER: the objective function has maximum value of 30 at (0,6)
Answer by mccravyedwin(408)  (Show Source):
You can put this solution on YOUR website!
What Greenestamps means is this:
The objective function z=x+5y can be written in slope intercept form y=mx+b
 where -1/5 is the slope, and z is the y intercept.
Think of all the parallel lines that have slope -1/5.
I'll draw a bunch of them across my graph:
You see that the only two of those parallel lines that intersect the feasible
region in only one corner points are the ones with y-intercept 6 and 0.
So the maximum value of z is 6 (when x=0, y=6) and the minimum value of z
is 0 (when x=0, y=0). Many times you can just visualize mentally how steep,
or how non-steep, the objective function slants, whether it slants upward
or downward and determine by inspection which are the points at which the
maximum and minimum values of z occur without substituting the corner points in
the objective function.
But be aware that it won't always be the top and bottom ones like it is in this
particular example. The shapes of feasible regions differ greatly.
Edwin
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