SOLUTION: A factory can produce two products, x and y, with a profit approximated by P = 14x + 22y � 900. The production of y can exceed x by no more than 200 units. Moreover, product

Question 665012: A factory can produce two products, x and y, with a profit approximated by P = 14x + 22y � 900. The production of y can exceed x by no more than 200 units. Moreover, production levels are limited by the formula x + 2y ≤ 1600. What production levels yield maximum profit?
A. x = 400; y = 600
B. x = 0; y = 0
C. x = 1,600; y = 0
D. x = 0; y = 200
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
Obvious constraints are:
and
(no negative production)
Given constraints are:

( cannot exceed by more than ,
the difference has to be or less),
and
.

Because of those constraints,
there is a feasibility region.
You only can work in that region of the x-y plane.
That region is bordered by the lines represented by
, , , and .
You can graph the lines.

graphs as the red slanted line ,
and graphs as the green line.
Your feasibility region is tue quadrilateral with
parts of the x-axis, the y-axis and the green and red lines for sides.
You can find the intersection points for each pair of lines.
For example, solving
gives you the solution with
for point (400,600), where the red and green slanted lines intersect.

The vertices of your feasibility region are:
(0,0) , (0,200) , (400,600) and (1600,0).
The maximum for will happen at 1 of those points.
(In some cases it could happen at 2 of vertices and the whole segment connecting them).
All you need to do is calculate for each of those 4 points.
I will show you the calculation for 2 of them:
For point (400,600), with and ,
.
For point (1600,0), .
The other points give you smaller values for ,
so the solution is point (1600,0),
with and .

THE REASON WHY IT WORKS THAT WAY:
The function to maximize,
is a function of x and y,
which could be represented in 3 dimensions,
wit being the third, dependent variable.
As with altitude as a function of 2-dimensional coordinates (latitude and longitude),
we can represent the function on paper by making a contour map.
The contour lines would be .
Luckily for us, the function is linear in and ,
so those contour lines will be straight lines, like the blue line below.

The blue line is the graph of
<--->
That is the line for .
As you change the constant, the line changes,
but all the other contour lines are parallel to that blue line.
As increase the value for the constant the lines moves away from the origin, until it moves out of your feasibility region.
You want the values (for , and for the largest possible ,
when you reach the end of the feasibility region.
In general, that will happen at one of the vertices, or at 2 of the vertices and the side that joins them.
In this case it happens at point (1600,0).
The maximum for P will be found at (1600,0)
where

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