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1. According to formula [19.12], Az* = Ab-> =0-0.1 + 1 -(-0.2) =

1. 4k* + 3u*2 = 3 > 2 and x* = 0; 5u* + lu\ = l and >•" = 3 > 0. Also 4x* + 5y* = 15 < 20 and u* = 0; 3x* + ly* = 21 and u\ = 1 > 0. So we see that [19.13] and [19.14] are satisfied.

3. (a) min 10.000,, + 8000,2 + n.000,3 s,. { ^ » ^

(b) Solution of the dual: max = 255.000 for ;tj = 100 and x2 = 450. Solution of the primal: min = 255,000 for (y,. y2, >'3) = (20.0,5).

(c) The minimum cost will increase by 2000.

5. (a) For x3 = 0, the solution is x\ = x2 = 1 /3. For x3 = 3. the solution is = 1 and = 2. (b) If 0 < < 7/3, then zm^(x3) = 2x3 + 5/3 for x\ = 1/3 and x2 = x3 + 1/3. If 7/3 < < 5, then zmax(*3) = x3 + 4 for x\ = - 2 and x2 = 5 - x3. If x3 > 5, then zm*x(x3) = 9 for x\ =3 and x2 = 0. (c) The solution to the original problem is X\ =3 and x2 = 0, with x3 as an arbitrary number > 5.

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