Problem $1$

Consider the below LP model:

Max Z $=100$x$1+80$x$2$

Subject to

x$1+$ x$2<=100$

$2$x$1+$ x$2<=160$

x$2<=50$

x$1,$ x$2>=0$

$1$a$)$ Find the optimal solution using the Graphical method. Make sure to use the objective function line

to determine the solution.

$1$b$)$ What is the objective value?

$1$c$)$ Specify the binding constraints of the model.

Problem $2$

Consider the same LP model as Problem $1.$

Now, write a $($different$)$ objective that results in a case of $$alternative optima$.$ Fully explain why this is the

case.

Problem $3$

Again, consider the same LP model as Problem $1,$ this time with the three $($main$)$ constraints converted

into $>=.$ What can you say about the optimal solution to this new model?

Problem $4$

One last time, consider Problem $1.$ Assuming that the optimal solution $($x$1*,$x$2*)$ stays the same as

Problem $1,$ recalculate the objective value $($without redoing the Graphical method$)$ when:

a$)$ The objective function changes to: Max $90$x$1+90$x$2$

b$)$ The objective function changes to: Max $150$x$1+60$x$2$