Question

Find necessary and sufficient conditions on the reals a and b under which the linear program

$\begin{array}{l}\mathbf{max}\mathbf{x}\mathbf{+}\mathbf{y}\\ \mathbf{ax}\mathbf{+}\mathbf{by}\mathbf{\le }\mathbf{1}\\ \mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\ge }\mathbf{0}\end{array}$

(a) Is infeasible.

(b) Is unbounded.

(c) Has a unique optimal solution.

Short answer

1. The solution is feasible in the origin$(x,y)=(0,0)$
2. The linear program is unbounded for $(a,b)\le 0$.
3. Conditions for optimal solution$a\ne b$ ,$a>0$ , and $b>0$.

Step-by-step solution

Explain Linear program.

Linear program is used for optimization tasks that has constraints and the optimization criterion as linear functions. A linear program has the set of variables that needs to be assign with the real values to satisfy the linear inequalities and to minimize or maximize a given linear objective function.

Conditions for Infeasible linear program

(a)

The given linear program is never infeasible for any value of $a,b$as there is always a feasible solution that satisfy $ax+by\le 1$. Also this solution can be feasible for origin $(x,y)=(0,0)$.

Conditions for unbounded linear program

(b)

This linear program is unbounded for$(a,b)\le 0$

This is because of the fact that if $a\le 0$, then increase the value of$x$ and same goes for $b$and$y$ . This will not violate our given constraints.

Conditions for a unique solution of linear program 

(c)

The finite optimal solution can be achieved if $a\ne b$, $a>0$, and$b>0$ .

If $a>b$, then the optimal solution can be achieved at $x=\frac{1}{b}$. Similarly, if $a<b$, then optimal can be found at $x=\frac{1}{a}$.

But if $a=b$ and$a,b$ are positive such that $x+y=\frac{1}{a}$achieve the optimum.

Hence, optimum exists and will be unique only if$a,b>0$ and$a$ is not equal to $b$.