Graphical solution (with two variables), solution properties.

Question

(c) infeasibility depends on the constraints; if we look at the graph we can see that (1, 1) is the intersection of constraint (I) and (II), and for this to be infeasible we need t in constraint (III) to $t>3$, because then if you won't find a point that satisfies all the constraints.

now the other three I'm not so sure, I know how to think about it graphically but I don't know how to do that algebraically. One optimal solution: if there is only one optimal solution that exists in a corner point of the feasible region. infinitely many optimal solutions: when the points lie on a line. unbounded: an open feasible region. I want to solve a, b, d algebraically.

A formal hint: $x_1, x_2 \leq 0$ is not a non-negativity condition. It has to be $x_1, x_2 \geq 0$. And your solution for a) is right. — callculus42, Aug 23 '15 at 21:05
a) is right? what do you mean by that? (Oh, maybe you meant (c) ) — Dick Armstrong, Aug 23 '15 at 21:11
I thought, that you commented a). If $t=3$, then all three constraints intersect at $(x_1,x_2)=(1,1)$. You can insert the values in the third constraint and calculate the value of t: $t \cdot 1+1= 4$ — callculus42, Aug 23 '15 at 21:17

callculus42 · Accepted Answer · 2015-08-24T03:52:23.017

Some hints:

a) If the slope of the third constraint not equal to the slope of the objective function and the slope varies between 3 and $-\frac{1}{2}$ then you have one optimal solution. Therefore $-\frac{1}{2} < t \leq 3$ and $t \neq 1$. If $t \leq -\frac{1}{2}$, then the third constraint is steeper (or equal slope) than the second constraint and the linear program becomes unbounded.

b) You have infinitely many solution if the line of the objective function can be (parallel) shifted onto a line of a constraint. That one line of a constraint has to have the same slope as the objective function. The objective function is $z=x_1+x_2$. Solving for $x_2$ gives $x_2=z-x_1$. Therefore the slope is $-1$. The third constraint has to have the same slope.

c) If you have no solution space, then the LP is not feasible. Therefore the slope has to be so small, that the third constraint line doesn´t intersect (1/1) any more. Thus the slope has to be greater than 3. The current solution space (without the third constraint) is the pink area.

d) t has to be smaller or equal $-\frac{1}{2}$. In this case the third constraint is steeper ( or parallel) than (to) the red line.

In general you should take a pencil, a ruler and a rubber and draw the third constraint for some values of t. For each value of t a new solution space has to be determined. Especially take care of the inequality signs.

from b) we get that $t=1$, and when $t=1$ we have infinitely many solutions, therefore in a) to get one optimal solution we know that t is NOT equal to 1, and like you said in the comments we can insert $(x_1, x_2)=(1,1)$ into the third constraint to get $t=3$, so we know that $1<t<=3$ and of course $t<1$ and now I'm going to try to figure out if t is greater than (or equal) to something when it's smaller than 1. — Dick Armstrong, Aug 23 '15 at 22:37
so, we already figured out that $1<t<=3$ and just now I got the following as well, $3<t<1$, by building a matrix out of constraint $(II)$ and $(III)$ and I just left the t as the first element in the second row and solved it by making the result (of either row) less than 1, i.e. $result<1$. Thank you so very much, I think that answered all of my questions, otherwise I'll be back =P — Dick Armstrong, Aug 23 '15 at 23:08
OH! and regarding c) and d), since $t=1$ gives infinitely many solutions, and $-3<t<1$ and $1<t<=3$ give only one optimal solutions, we know that we have no solution when $3>t$ and unbounded when $t<=-3$ — Dick Armstrong, Aug 23 '15 at 23:17

Graphical solution (with two variables), solution properties.

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