Implement the following $LP$ problem $in$ a spreadsheet. Use Solver $to$ solve the problem and create a Sensitivity Report. Use this

information $to$ answer the following questions:

MAX: $4x_1+4x_2$

Subject $to$: $2x_1+4x_{2}20$

$3x_1+5x_{2}15$

$x_1,x_{2}0$

$(a)$ What range $of$ values can the objective function coefficient for variable $x_1.x_1$ can decrease $by$

$or$ increase $by$

without changing the optimal solution.

$(b)Is$ the optimal solution $to$ this problem unique, $or$ are there alternate optimal solutions?

None $of$ the allowable increase $or$ decrease values for the objective coefficients are zero, $so$ the optimal solution

$is$ unique.

$We$ cannot determine $if$ the optimal solution $is$ unique based $on$ our $sensitivity$ report because the solution $is$

degenerate.

None $of$ the allowable increase $or$ decrease values for the $RHS$ values are zero, $so$ the optimal solution $is$

unique.

Some $of$ the allowable increase $or$ decrease values for the $RHS$ values are zero, $so$ there are alternate optimal

solutions.

Some $of$ the allowable increase $or$ decrease values for the objective coefficients are zero, $so$ there are alternate

optimal solutions.

$(c)$ How much does the objective function coefficient for variable $x_2$ have $to$ increase before $it$ enters the optimal solution $ata$

strictly positive level? $(Round$ your answer $to$ three decimal places.$)$

$(d)$ What $is$ the optimal objective function value $if{x}_2$ equals $1?(Round$ your answer $to$ three decimal places.$)$

$(e)$ What $is$ the optimal objective function value $if$ the $RHS$ value for the second constraint changes from $15to20?(Round$ your

answer $to$ three decimal places.$)$

$(f)Is$ the current solution still optimal $if$ the coefficient for $x_{2}in$ the second constraint changes from $5to1?$ Explain. $(Round$

your answer $to$ three decimal places.$)$

$Ifwe$ change this coefficient from $5to1,$ then the new reduced cost for $x_2$ for our current solution would $be$

$.$ Therefore, $itbe$ profitable $to$ increase the value $of{x}_2$ and the current solution would

$be$ optimal.