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A year ago
 A company manufactures 50-inch and 75-inch rear projection television sets. Each 50-inch set ... A company manufactures 50-inch and 75-inch rear projection television sets. Each 50-inch set contributes 200 to profits and each 75-inch set contributes 475 to profits. The company has purchase commitments for 500 50-inch sets and 200 75-inch sets for the next month, so they want to make at least that many television sets. Although they think they can sell all the 50-inch sets that they could currently make, they do not think they can sell more than 375 75-inch sets. Their production capacity allows them to make only 975 sets in total including both types of television sets. They want to know how many of each type to make so as to maxim-ize profits.  a. What is the objective function for this LP problem?  b. What are the constraints involving x1, assuming that x1 corresponds to 50-inch TV sets?  c. What is the optimal solution point for this problem?  d. What is the optimal value of the objective function? Read 20 times 1 Reply
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A year ago
 a. The objective function is Max z = 200x1 + 475x2.x1 = number of 50-inch TV sets made; x2= number of 75-inch TV sets madeMax Z = 200x1 + 475 x2subject to x1  500 (commitments)x2  200 (commitments)x1 < 375 (limits on sales)x1 + x2  975 (capacity)b. From the values of commitments and capacity given, the constraints involving x1, assuming that x1 corresponds to 50-inch TV sets, is x1  500; x1 + x2  975.c. The optimal solution point is x1 = 600; x2 = 375.x1 = 500 and x2 = 200 => (500, 200)x1 = 500 and x2 = 375 => (500, 375)x1 = 375 and x1 + x2= 975 => x1 = 600 => (600, 375)x2= 200 and x1 + x2 = 975 => x1= 775 => (775, 200)d. The optimal value of objective function is 298,125.Max z = 200x1 + 475x2= 200(500) + 475(200) = 195,000= 200(500) + 475(375) = 278,125= 200(600) + 475(375) = 298,125 optimum= 200(775) + 475(200) = 250,000
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