Consider the following linear program.

Min $8x+12Y$

$s.t.$

$1x+3y6$

$2x+2Y8$

$6x+2y12$

$x,y0$

$($a$)$ Use the graphical solution procedure to find the optimal solution.

What is the value of the objective function at the optimal solution?

$at(x,)=(,)$

$($b$)$ Assume that the objective function coefficient for $x$ changes from $8$ to $6.$ Use the graphical solution procedure to find the new optimal solution. Does the optimal solution change?

The extreme point $(x,y)=(,)$ remains $v$ optimal. The value $of$ the objective function becomes

Does the optimal solution change?

$($d$)$ The computer solution for the linear program in part $($a$)$ provides the following objective coefficient range information.

$\backslash$table$[[$Variable$,\backslash$table$[[$Objective$],[$Coefficient$]],\backslash$table$[[$Allowable$],[$Increase$]],\backslash$table$[[$Allowable$],[$Decrease$]]],[x,8\mathrm{.}00000,4\mathrm{.}00000,4\mathrm{.}00000],[Y,12\mathrm{.}00000,12\mathrm{.}00000,4\mathrm{.}00000]]$

How would this objective coefficient range information help you answer parts $($b$)$ and $($c$)$ prior to re$-$solving the problem?

The objective coefficient range for variable $x$ is $$ to $.$ Since the change in part $($b$)$ is $-$ Select $$ this range, we know the optimal solution $-$ Select $-vv$ change. The objective coefficient range for variable $Y$ is Since the change in part $($c$)$ is $-$ Select$-v$ this range, we know the optimal solution $-$ Select$--v$ change.