Question: Given the following answer and sensitivity reports Cell SE$12 Objective Cell (Max) Name profit per board Total Profit Original Value $0.00 Final Value $4,320.00 Variable

Given the following answer and sensitivity reports Cell SE$12 Objective Cell (Max) Name profit per board Total Profit Original Value $0.00 Final Value $4,320.00 Variable Cells Name boards X boards Y boards z Original Value 0 Cell SB$9 SC$9 $D$9 1 Final Value 0 0 720 Integer Contin Contin Contin 0 0 Cell $E$15 SES16 $E$17 Constraints Name Celmacs time Lhs Test stand time Lhs Graphic cards demand Lhs Cell Value 3600 1440 2160 Formula Status Slack $E$15=SG$17 Not Binding 2060 Variable Cells Objective Coefficient 2 Cell $B$9 $C$9 SD$9 Final Reduced Value Cost 0 -0.4 0 -2.8 720 0 Name boards X boards Y boards z Allowable Increase 0.4 2.8 1E+30 Allowable Decrease 1E+30 1E+30 8 6 Constraints Cell SE$15 $E$16 $E$17 Name Celmacs time Lhs Test stand time Lhs Graphic cards demand Lhs Final Value 3600 1440 2160 Shadow Price 1.2 0 Constraint Allowable RH. Side Increase 3600 275 1550 1E+30 100 2060 Allowable Decrease 3433.333333 110 1E+30 0 3600 275 SES15 SES16 1.2 0 3600 1550 3433.333333 110 1440 1E+30 Celmacs time Lhs Test stand time Lhs Graphic cards demand Lhs SES17 2160 0 100 2060 1E+30 The optimal solutions are X= Y= and Z= The optimal profit is Constraint (1,2,3?) is working at full capacity. The unused capacity of the third constraint is If we add 10 units capacity to constraint 1 at a cost of $1 per unit, the amount of net benefit to the objective function is dollars

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