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Lecture on Chapter 3 The Linear Programming Begins...

This review gives a brief summary of the graphical method.

Inequalities An inequality in one variable looks like
                                             $2 x \leq 2$
and is solved by rearranging it so only the variable appears on the left hand side:
                                                   $x \leq 1$ .
This can also be done graphically as follows:

Convert it to an equation and solve it:
                                            $2 x + 3 = 5$
                                    leads to $x = 1$ as a solution.

On the number line,
                                    plot the point $x = 1$
and notice that all points to the left of it satisfy the inequality
and the ones on the right don't.
Simply try values like $x = 0$ and $x = 2$ .
Thus the answer is the interval ( $- ∞, 1$ ]. In picture,
Similarly, if we have another inequality, say
$0 \leq 3 x + 6$ ,
then we again solve the equation
$3 x + 10 = 4$ and get $x = -2$ .
We conclude that points on the right side of this solution satisfy the inequality,
giving the interval [-2, $∞$ ) .

In picture,
If we combine the above two inequalities and ask for a solution,
then we get the intersection of the two intervals which is
[ $-2, 1$ ].
In pictures:

We can describe this process as follows.
Convert both inequalities to equations and solve each separately .
Thus, we get $x = 1$ and $x = -2$ . Plot these points on the number line to get the following picture.

This gives three intervals ( $- ∞, -2$ ),( $-2, 1$ ),( $1, ∞$ ). We test each at some convenient point to check if all the inequalities are satisfied and deduce the answer [ $-2, 1$ ].
We wish to generalize this to two or more varaibles.
Inequalities in two variables An inequality like
                                                  $x + 2 y \leq 4$
is the next topic. As before, we first convert it to the equation
                                                 $x + 2 y = 4$ .
We note that this is a line and we know how to plot it. It is not difficult to see that the plane is split into two halves so that on one side of the line the inequality is true, while on the other side it is not!
Thus,
                                      having plotted $x + 2 y = 4$
we see that
                                 at $x = 0, y = 0$ the inequality is satisfied |
and so the solution for the inequality is the half plane containing the origin!
(We show the part in the first quadrant only).
If we have more than one inequalities, then we solve them separately and take the common part. Here is the solution for
$0 \leq x, 0 \leq y, 3 x + 4 y \leq 12, 2 \leq x + 2 y$ .

See if you can figure it out for yourself.
(The second region extends to infinity, but we have drawn a small portion!)
The blue region is the final answer.

Step 1. $0 \leq x, 0 \leq y, 3 x + 4 y \leq 12$ .

Step 2. $0 \leq x, 0 \leq y, 2 \leq x + 2 y$ Actually, the green region extends out to infinity!
Step 3. Intersect the above two regions! The blue part is the answer.
General Method On a common graph paper, draw the equations corresponding to each inequality, and
mark the regions indicated by each inequality using a directional arrow.
In the above pictures, we have not marked the arrows on the two axes, since we usually work in the first quadrant anyway!

Take the common part.
The result will be a polygon, which may have open regions running off to infinity in some directions; such regions are unbounded.

Actually, the polygons that you can get this way are convex,
meaning if two points are inside such a polygon then the line segment joining them is also inside.
This is a very important property and regions satisfying this property are important objects of study!

Can you think of an example of a polygon which will not have this property?

Next calculate all the corner points by finding common points of the equations, two at a time.
Note that some of the common points may be outside the final region (in the above example, the point ( $0, 0$ ) and the point (8,-3) are the "lost" points.
Sometimes, entire lines may get lost.
In the above work, we did not even bother with the line $y = 12$ since it was outside our yellow triangle and the desired inequality $y \leq 12$ covered it completely.
Be on the lookout for such spurious inequalities! They might be meaningful for the practical proble but may not have a role to play in the solution!
Practical setup Be sure to review various examples in the section 3.2 where a practical situation leads to a set of inequalities and a linear function to be optimized (i.e. maximized or minimized).
The first step is to clearly name the variables and write down the inequalities and the function to be optimized.
Then you sketch the region carefully, provided that you have only two varaibles. For more variables, you need to go to the next chapter and switch over to algebra; since the geometry gets too complicated!
Then you list all the corner points of the region in order and evaluate the function at each of the corners.
The optimum value of the function is among the values at the corner points, with one exception: if the region is not bounded, then the function may not have a maximum or a minimum. This can be decided by checking along the edges of the polygon running off to infinity.

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Explanation

Why does the graphical method work?

Parametric lines. Consider a line in the plane, say:
$y = 3 x + 5$
It passes through a point ( $1, 8$ ). We want to study the line near this point.
So we take a point on this line whose x-coordinate is $1 + t$ and notice that its corresponding y-coordinate shall be $8 + 3 t = 8 + 3 t$ .
In fact, all points of this line can now be described by the parametric equations:
$x = 1 + t, y = 8 + 3 t$
This is called the parametric form of the line.
It is useful to be able to calculate such a form for any line near any convenient point!
Thus for the same line, we could also have started with a point ( $-2, -1$ ) and
concluded a different parametric form
$x = -2 + t, y = -1 + 3 t$
Functions on Parametric lines. Consider a function, say $f (x, y) = 3 x + 4 y$ . We can analyze how it behaves on our parametric line by plugging in the parametric form:
$f (x, y) = 35 + 15 t$ or $f (x, y) = 35 + 15 t$ . This shows that as $t$ increases, so does the function value. Remember that the parametr $t$ was the change in the x-coordinate from the point ( $1, 8$ ).
Thus, as we let x-coordinate increase on our line, the function value increases.

Conclusion.

Thus on a line in parametric form, a linear function increases or decreases with the parameter depending on the coefficient of the parameter.
For the above line, if we consider a different function, say

$g (x, y) = 3 x - y + 2$ then we see that

$g (1 + t, 8 + 3 t) = -3$ and this simplifies to

$g (1 + t, 8 + 3 t) = -3$ .

Thus, a linear function on a line is either constant at all points or increases in one of the two available directions and decreases in the opposite direction.

Consider the plane region that we can plot for our problem. Where would a linear function become maximum on such a region. If we take any point in the interior of our region we can draw a little line segment through the point which is still entirely in the region. Now if our function is not constant on the line, it would be increasing in one of the two directions and thus would not be maximum at our given point.
If by luck, we had chosen a line segment on which the function happens to be constant, we can choose a different segment through the same point and make the same argument. We could not have the same function constant on the second segment as well, for it is clear that then the function would be identically constant on the whole plane and our problem has a trivial answer: every point is a maximum point!

Thus, our maximum point has to be on the boundary! It could be on a boundary segment or at a corner. Note that if it is on a segment, but not at a corner, then the function has to be constant on the whole boundary segment (for otherwise we get a contradiction as above). Thus, it would also be maximum at one of the endpoints which must be a corner.

This is why it is enough to only check the corner points for locating a maximum.

This also explains that if we find two maximum points which are corners, then the line joining them must form a boundary line, i.e. they must be adjacent points on the boundary polygon.

If the region is unbounded, then it has boundary lines running off to infinity and we may find that the maximum point may not exist in the sense that it has to be a point of infinity on one such boundary line.

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Problem 3.2.2 by graphing

We discuss the solution of the following problem:
Maximize the profit
$P = 2 x + 1.5 y$
subject to
$0 \leq x, 0 \leq y, 3 x + 4 y \leq 1000$ and $6 x + 3 y \leq 1200$ .
We first sketch the lines and find their common point.
Then we decide on the region.
Note that in the picture below, the inequality $3 x + 4 y \leq 1000$ corresponds to the line BC
and the inequality $6 x + 3 y \leq 1200$ matches the line AB.
Their regions both point towards the origin, since the origin satisfies both of them!
The axes are automatically included with regions pointing towards the first quadrant.

Intersect the lines $3 x + 4 y = 1000$ and $6 x + 3 y = 1200$ .

> 3*x+4*y <= 1000;

$3 x + 4 y \leq 1000$

> solve({3*x+4*y=1000,6*x+3*y=1200},{x,y});

$\{x = 120, y = 160\}$

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> P1:=textplot([3,3,`O`],color=blue,font=[TIMES,BOLD,16],align={ABOVE,RIGHT}):
P2:=textplot([200,0,`A`],color=blue,font=[TIMES,BOLD,16],align={ABOVE,RIGHT}):
P3:=textplot([120,160,`B`],color=blue,font=[TIMES,BOLD,16],align={ABOVE,RIGHT}):
P4:=textplot([0,250,`C`],color=blue,font=[TIMES,BOLD,16],align={ABOVE,RIGHT}):
l1:=plot((1000-3*x)/4,x=0..150,color=red,thickness=3):
l2:=plot((1200-6*x)/3,x=110..210,color=blue,thickness=3):
l3:=line([0,0],[200,0],color=black,thickness=3):
l4:=line([0,0],[0,250],color=black,thickness=3):
region:=polygonplot([[0,0],[200,0],[120,160],[0,250]],color=yellow):
display({P1,P2,P3,P4,l1,l2,l3,l4,region});

Now we list all the corners and evaluate the function $P = 2 x + 1.5 y$ at each corner.

> corners:=[[0,0],[200,0],[120,160],[0,250]];

$corners := [[0, 0], [200, 0], [120, 160], [0, 250]]$

> funvals:=[seq(2*corners[i][1]+(1.5)*corners[i][2],i=1..nops(corners))];

$funvals := [0., 400., 480.0, 375.0]$

This shows that the max. occurs at the third point $[120, 160]$ and the max. value is 480.
Here is an animated display that shows why this works!

>    with(plottools):
F := proc(t)
  plots[display](l1,l2,region,
     line([t/2,0],[-t/4,t],color=magenta,thickness=2));
     end:
animate(F,[t],t=150..480,
                scaling=constrained,axes=none);

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Some B2 like problems.

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Code

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Scratch