Maximum value(LPP)

Question

Let $\alpha \in R $ . If $(3,0,0,\beta)$ is an optimal solution of LPP,

Minimize $x+y+z-\alpha t$

s.t

$ 2x-y+z=6$

$-x+y+t=3$

$x,y,z,t \ge 0 $

The the maximum value of $\beta - \alpha$ is ?

My attempt: I tried by simply plugging in the value of optimal solution in constraint equation(2nd one) and got the value of $\beta = 6$. Now to find the value of $\alpha$, I put the Min= - Max That is the problem is reduced to

-Maximize $-x-y-z+ \alpha t$

Also, it can be observed that z and t seems to be slack variables for this problem. Hence, as the problem is not Maximum type but of -Maximum type, therefore slack variables must be subtracted from the -maximization equation instead of adding.(Is this correct?)

Therefore to make the above equation as -Maximize $-x-y-z-t$ we have to put $\alpha = -1$ So, the maximum value must be 6-(-1)=7

Answer 1

You have arrived at the right numerical answer, but, to be honest, I don't follow your logic in the last two paragraphs.

Also, it can be observed that z and t seems to be slack variables for this problem. Hence, as the problem is not Maximum type but of -Maximum type, therefore slack variables must be subtracted from the -maximization equation instead of adding.(Is this correct?)

1. Rewriting the objective function doesn't change the constraint.
2. It's possible that a max problem contains a '≥' constraint, even though it's "bizarre" in the sense of the "SOB" method.
3. There's no such thing callled "maximization equation". The function to maximize is called the "objective function, while the constraints in the given LPP are all equations.

Therefore to make the above equation as -Maximize $-x-y-z-t$ we have to put $\alpha=-1$. So, the maximum value must be $6-(-1)=7$.

I don't follow your idea as I'm already lost in the previous paragraph.

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I agree with your observation that $z$ and $t$ "seem" to be a slack variable. To put this more concretely, transform the given LPP into a new LPP having only two decision variables $x$ and $y$. To make things simple, I keep the LPP type (minimization) unchanged throughout my answer.

1. Make each of $z$ and $t$ the subject in the corresponding equality constraint, then substitute them into the objective function in the given LPP, so that it becomes $\text{constant} -(\alpha+1)x - (\alpha-2)y$. The constant term can be neglected.

2. Use the "slackness" $z \ge 0$ and $t \ge 0$ to get two inequality constraints $2x-y \le 6$ and $-x+y \le 3$.

3. (Optional) Plot the feasible region of the transformed LPP.

   

4. Invoke the Fundamental Theorem of LP to conclude that an optimal solution is attained at one of the extreme points of the feasible region:

   coordinates	objective function value	optimal solution?
   $(0,0)$	$0$	maybe
   $(0,3)$	$-3\alpha+6$	no
   $(3,0)$	$-3\alpha-3$	sure
   $(9,12)$	$-21\alpha+15$	maybe

5. Deduce from the given optimal solution that $\alpha \ge -1$ and $\alpha \le 1$.

6. The previous step justifies the choice of $\alpha = -1$ for calculation the maximum possible value of $\beta - \alpha$.