this is for linear programming
6-2 Eli Orchid can manufacture its newest pharmacutical product in any...
6-2 Eli Orchid can manufacture its newest pharmacutical product in any of three processes. One costs $14,000 per batch, requires 3 tons of one major ingredient and 1 ton of the other, and yields 2 tons of output product. The second process costs $30,000 per batch, requires 2 and 7 tons of the ingredients, respectively, and yields 5 tons of product. The third process costs $11,000 per batch, requires 9 and 2 tons of the ingredients, respectively, and yields 1 ton of product. Orchid wants to find the least costly way to produce at least 50 tons of the new product given that there are 75 tons of ingredient 1 and 60 tons of ingredient 2 on hand. (An optimization output appears in Table 6.70.) (a) Briefly explain how this problem can be modeled by the LP min 140, +30+2 + 1113 s.t. 23, +52, + 110 > 50 3.0, +222 +913 < 75 12, + 72, +213 < 60 I1, I2, I3 > 0 (b) Coef through (d) as in Exercise 6-10 Table 6.7 Optimization Output for Eli Orchid Exercise 6-2 Solution value (min) = 311.111 VARIABLE SENSITIVITY ANALYSIS: Name Optimal Bas Lower Upper Object Value Bound Bound x1 5.556 BAS 0.000 14.000 7.778 BAS 0.000 + Tin 30.000 0.000 NBL 0.000 | virlin 11.000 CONSTRAINT SENSITIVITY ANALYSIS: Name Typ OptimalRHS Slack Lower Dual Coer Range 7.556 50.000 -0.000 42.857 0.000 75.000 42.778 32.222 -1.111 60.000 0.000 25.000 Reduced Object 0.000 0.000 5,667 Lower Range 12.000 -infin 5,333 Upper Range vintina 35.000 vintin Upper Range 70.263 infin 70.000 Identify the resource associated with the objective function and each main constraint in part (c) Identify the activity associated with each decision variable in part (a). Interpret the left-hand-side coefficients of each decision variable in part (a) as inputs and outputs of resources per unit activity.