For the following primal problem,

(a) formulate the dual,

(b) solve the dual graphically. Then use the dual solution to find the optimal values of

(c) the primal objective function and

(d) the primal decision variables.

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Minimize subject to c = 36y1 +30y2 + 20y3 6y13y2 y3 10 2y1 +5y2+4y3 8 Y1, y2 y30 (a) The dual is Maximize subject to = 10x1 +8x2 6x1 + 2x2 36 3x1 +5x230 x1 + 4x2 20 X1, X20 The dual was solved graphically as a primal in Problem 7.10, where = 5, x2 = 3, and 7 = 74. (b) (c) With 74, 7 = 74. (d) To find the primal decision variables y, y2, y3 from the dual decision variables x1, x2, first convert the primal (I) and dual (II) inequalities to equations. I. 6y1+3y2 y3 - s = 10 2y1+5y2+4y3 - 82 = 8 II. 6x+2x2 +11=36 3x15x2+12 = 30 x1 + 4x2 + 13 = 20 Then substitute = 5, x2 = 3 into the dual constraint equations to solve for the slack variables 11, 12, 13. 6(5)+2(3) +1 = 36 (7 = 0) 3(5)+5(3) +12 = 30 (12 = 0) (5)+4(3) +13 20 (73 = 3) With 73 0, the corresponding decision variable 3 must equal zero. With = 72 = 0, 1, 2 0. Since the optimal values of the dual decision variables X1, X2 do not equal zero, the corresponding primal surplus variables 51, 52 must equal zero. Incorporating the knowledge that y3 = 51 52 = 0 into the primal constraint equations, 6y13y2 (0) - (0) = 10 2y15y2+4(0) - (0) = 8 Solved simultaneously, i = 1.08 and 2 = 70 1.17. Thus, the primal decision variables, allowing for rounding, are y = 1.08, y2 = 1.17, and 3 = 0.