SOLUTION: use linear programming to maximize z=10x+15y subject to x+4y < 360 2x+y<300 x>0, y>0 all signs are equal to

Question 1044529: use linear programming to maximize z=10x+15y subject to
x+4y < 360
2x+y<300
x>0, y>0
all signs are equal to
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
The constraints are
. That is the quadrilateral limited by the lines : .
Sketching a graph is very helpful.
It allows you to "see" the feasible region you are working with.
In this case, it is the region below both blue an green lines,
above the x-axis, and
to the right of the y-axis.
The feasible region is obviously the quadrilateral .
It is easy to see that shrewdly selected test point ,
within , meets the requirement of all the constraints,
and if you move away from that point crossing over any boundary line,
you cease to meet all requirements.

The line is the y-axis.
The line is the x-axis.
Those two lines intersect at , the origin.

To graph , I connected with a straight line its x- and y-intercept.
I found those intercepts by solving
--->---> to get point ,
and
---> to get point .

To graph , I did the same.
I found its x- and y-intercepts by
solving
--->
---> to get point ,
and
--->
---> to get point .

I also had to find the point where and intersect.
I did it by solving
to get point .
There are many ways to salve that. Here is one:
-->-->-->-->-->-->-->-->-->-->

Finally, we need to maximize , a function of and , within that feasible region.
Because the boundaries and the function are linear,
the maximum will be
either at one of the vertices of the feasible region,
or all along one of the edges of that region.

It could have been obvious to us from the start that the maximum would be at point (see note below),
but the recipe usually taught to solve this kind of problems involves calculating the value of the function at every vertex.
The idea may be the educational value of
encouraging practice,
valuing hard work, and
helping you visualize the situation.
So, we worked hard to get to this point, and we continue along the expected path.

At :
, obviously no maximum.
At :

At :

At :
.

The maximum is at point .

NOTE:
Versions of this type of problem are very popular.
You have , telling you the feasible region is in quadrant I.
There are two boundary line constraints: ,
with all positive coefficients,
telling you that the feasible region is below two lines with positive x- and y-intercepts.
and the function to maximize, , has positive coefficients.
To make you work harder, hey make the maximum be at the intersection of the two lines,
by making the slope of the lines , ,
be in between the slopes, and , of the two slanted boundary lines.
So, in this case, we could immediately see that
the boundary line slopes are and ,
with the slope of any line,
, in between them .
That tells us that the maximum is at the intersection of the two slanted boundary lines,
---> point .
So the maximum is .
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