X tons of steel i produced max z= (51-30-8-5)x1+(30-20-5)x2+(25-10-8-5)x3 st Optimal solution found. Objective: EQUATION NAME...

 

 

 

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X₁ tons of steel i produced max z= (51-30-8-5)x1+(30-20-5)x2+(25-10-8-5)x3 st Optimal solution found. Objective: EQUATION NAME 4. Steelco uses coal, iron and labor to produce 3 types of steel. The inputs for one ton of each type of steel are shown in the following table. Up to 200 tons of coal can be purchased at a price of $10 per ton. Up to 60 tons of iron can be purchased at $8 per ton and up to 100 labor hours can be purchased at $5 per hour. amac const (1) const (2) const (3) VARIABLE NAME 3x1+2x2+x3<200 x1+x3<60 x1+x2+x3<100 Answer the below questions according to the gams report given below. a. What would be the profit if only 40 tons of iron could be purchased? b. What would be the profit if 230 tons of coal could be purchased? c. What is the smallest price per ton for steel 3 that would make it desirable to produce it? d. What would be the profit if steel 1 sold for $55 per ton? e. What would be the profit if steel 2 sold for $28 per ton? f. What would be the profit if steel 1 sold for $48 per ton and steel 2 sold for $28 per ton? g. What would be the new optimal solution if 230 tons of coal and 40 tons of iron are available? x (1) x (2) x (3) Z 1 2 3 1 IN M 2 3 steel 1 2 3 EQU amac EQU const LOWER -INF -INF -INF VAR X LOWER coal 3 ton 2 ton 1 ton LEVEL 530.000000 LEVEL 60.000 10.000 LOWER UPPER 200.000 200.000 60.000 60.000 70.000 100.000 UPPER Iron 1 ton 0 ton 1 ton +INF +INF LEVEL MARGINAL 2.500 0.500 MARGINAL -1.000 LOWER -INF 180 UPPER LOWER 0 70 labor 1 hour 1 hour 1 hour 7.5 0 -INF -INF Selling price $51 $ 30 $25 CURRENT MARGINAL 1.000 0 200 60 100 CURRENT 8 5 2 1 UPPER +INF 260 66.67 +INF UPPER +INF 5.333 3 +INF || X₁ tons of steel i produced max z= (51-30-8-5)x1+(30-20-5)x2+(25-10-8-5)x3 st Optimal solution found. Objective: EQUATION NAME 4. Steelco uses coal, iron and labor to produce 3 types of steel. The inputs for one ton of each type of steel are shown in the following table. Up to 200 tons of coal can be purchased at a price of $10 per ton. Up to 60 tons of iron can be purchased at $8 per ton and up to 100 labor hours can be purchased at $5 per hour. amac const (1) const (2) const (3) VARIABLE NAME 3x1+2x2+x3<200 x1+x3<60 x1+x2+x3<100 Answer the below questions according to the gams report given below. a. What would be the profit if only 40 tons of iron could be purchased? b. What would be the profit if 230 tons of coal could be purchased? c. What is the smallest price per ton for steel 3 that would make it desirable to produce it? d. What would be the profit if steel 1 sold for $55 per ton? e. What would be the profit if steel 2 sold for $28 per ton? f. What would be the profit if steel 1 sold for $48 per ton and steel 2 sold for $28 per ton? g. What would be the new optimal solution if 230 tons of coal and 40 tons of iron are available? x (1) x (2) x (3) Z 1 2 3 1 IN M 2 3 steel 1 2 3 EQU amac EQU const LOWER -INF -INF -INF VAR X LOWER coal 3 ton 2 ton 1 ton LEVEL 530.000000 LEVEL 60.000 10.000 LOWER UPPER 200.000 200.000 60.000 60.000 70.000 100.000 UPPER Iron 1 ton 0 ton 1 ton +INF +INF LEVEL MARGINAL 2.500 0.500 MARGINAL -1.000 LOWER -INF 180 UPPER LOWER 0 70 labor 1 hour 1 hour 1 hour 7.5 0 -INF -INF Selling price $51 $ 30 $25 CURRENT MARGINAL 1.000 0 200 60 100 CURRENT 8 5 2 1 UPPER +INF 260 66.67 +INF UPPER +INF 5.333 3 +INF ||

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The image youve provided appears to be a linear programming problem that involves the production of three types of steel using resources such as coal ... View the full answer