Redo Problem 8.4 for the following primal problem: Maximize subject to (a) The dual is Minimize subject

Redo Problem 8.4 for the following primal problem:

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Maximize subject to (a) The dual is Minimize subject to = 36x+84x2 + 16x3 3x16x2+x3 20 2x16x2+4x3 15 X1, X2, X30 c = 20y1 + 15 y2 3y1 +2y2 36 y +4y2 16 6y1 + 6y2 84 y1 y20 (b) The dual was solved graphically as a primal in Problem 7.14, where y = 8, 2 = 6, and 7 = 250. (c) With 250, = 250. = (d) I. 3x+6x2 + x3 +$1 = 20 II. 3y+2y2 11 = 36 2x1 + 6x2+4x3 + S2 = 15 6y16y2 12 = 84 y1 +4y2 13 16 Substituting = 8 and y = 6 in II, 3(8)+2(6) 11 = 36 (7 = 0) 6(8)+6(6) 12 = 84 (12 = 0) (8)+4(6) 13 = 16 (73 = 16) With 71 =12=0,x1, x2 0. Since 130, x3 = 0. With 1, 20, 51 = 52 = 0. Substituting in I, 3x16x2+(0) + (0) = 20 2x1+6x2+4(0) + (0)=15 Solved simultaneously, = 5 and 2 = {. Thus, = 5, = {, and x3 = 0.