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Unformatted text preview: Quiz Sample Solution Due: February 24, 2009 Problem 1 [20 points] Top Brass Trophy Company makes large championship trophies for youth athletic leagues. They are planning production for fall sports:football and soccer. Each football trophy has a wood base, engraved plaque, and a large brass football on a top, and returns $12. Soccer trophies are similar except that a brass soccer ball is on a top and the unit profit is $9. The football base requires 4-feet board, while the soccer base requires 2-feet board. The company has 1000 brass footballs in stock, 1500 soccer brass balls, 1750 plaques, and 4800 feet of board. Assume that what is produced can be sold. (a) [10 points] Formulate a linear programming model corresponding to maximizing the companys profit subject to the given resource constraints. (b) [10 points] Write the problem in the standard form Solution . (a) Let x 1 be the number of football trophies and x 2 be the number of soccer trophies to be produced. The profit is $12 and $ 9 per football and soccer trophy, respectively. The objective is to maximize 12 x 1 + 9 x 2 . The board requirements of 4 feet and 2 feet per football and soccer trophy, respectively, and the available resources of 4800 board feet yield the following constraint 4 x 1 + 2 x 2 4800 . Since both trophies use the same plaque, whose total availability is 1750, we have x 1 + x 2 1750 . Since each trophy requires its own brass ball, based on the available 1000 brass football balls and 1500 brass soccer balls, we have the following two constraints x 1 1000 , x 2 1500 . 1 We also have x 1 0 and x 2 0. Thus, the LP model is: maximize z = 12 x 1 + 9 x 2 subject to 4 x 1 + 2 x 2 4800 x 1 + x 2 1750 x 1 1000 x 2 1500 x 1 , x 2 . (b) To bring the problem in the standard form, we add slack variables in all four inequality constraints (aside from the sign constraints on the variables), and we require slack variables to be nonnegative. Thus, the problem in the standard form is: maximize z = 12 x 1 + 9 x 2 subject to 4 x 1 + 2 x 2 + s 1 = 4800 x 1 + x 2 + s 2 = 1750 x 1 + s 3 = 1000 x 2 + s 4 = 1500 x 1 , x 2 , s 1 , s 2 , s 3 , s 4 . Problem 2 [25 points] Solve the following problem by M-method or by the two- phase method....
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- Spring '09