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Unformatted text preview: For a football, 4/5 = 12/ c 2 4 c 2 = 60 c 2 = 15 Thus, if the profit for a football decreased to $15 or less, point B will also be optimal. The solution at B is x 1 = 128.5, x 2 = 57.2, and Z = $2,400. (c) If the constraint line for rubber changes to 3 x 1 + 2 x 2 = 1,000, it moves outward, eliminating points B and C . However, since A is the optimal point, it will not change and the optimal solution remains the same, x 1 = 0, x 2 = 160, and Z = 2,560. There will be an increase in slack, s 1, to 680 lbs. If the constraint line for leather changes to 4 x 1 + 5 x 2 = 1,300, point A will move to a new location, x 1 = 0, x 2 = 250, Z = $4,000....
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This note was uploaded on 02/02/2012 for the course EBA 3334 taught by Professor Faruque during the Fall '11 term at FIT.
 Fall '11
 faruque

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