Engineering Optimization: Theory and Practice, Fourth Edition

(Martin Jones) #1
Answers to Selected Problems 797

3.55xA∗= 1. 5 ,xB∗= 0 3 .57x∗m= 61 , xd∗= 02


3.60x∗= 63 / 11 , y∗= 53 / 11


3.66all points on the line joining (7.4286, 15.4286) and (10, 18)


3.71x∗= 3. 6207 , y∗= 8. 4483 3.75x∗= 2 / 7 ,y∗= 03 / 7


3.79x∗= 65 / 23 , y∗= 54 / 23 3.85x∗= − 4 / 3 ,y∗= 7


3 .89x∗= 0 ,y∗= 3


3 .92(x 1 , x 2 )=amounts of mixed nuts (A, B) used, lb.x∗ 1 = 08 / 7 , x 2 ∗= 201 / 7


3.94xA∗= 26. 5 , xB∗= 13. 25


3.96xi= umber of units ofn Piproduced per week.x 1 ∗= 001 / 3 , x∗ 2 = 502 / 3


3.99(x 1 , x 2 )=number of units of (A, B) sold per month.x 1 ∗ 9 = 1. 17 , x∗ 2 = 54


3.102xi= umber of days used in a month for process typen i (i= 1 , 2 , 3 , 4 ).
x 1 ∗= 03 , x∗ 2 =x 3 ∗=x 4 ∗= 0


CHAPTER 4


4.1 X∗= { 2. 333 , 1. 333 , 0 , 0 }

4.3xi∗= 0 ,i= 1 , 2 , 3 , x∗ 4 = 2 / 5 , x∗ 5 = 4 / 5 4 .5solution unbounded


4.9xi∗= 0 ,i= 1 , 2 , 5 , 6 , 7 , x 3 ∗= 0. 5 ,x 4 ∗= 1. 5


4 .12x 1 ∗= 2. 35 , x 2 ∗= 0. 1 , x 3 ∗= 2. 7 , x∗ 4 = 1. 2


4 .15x 1 ∗=x 2 ∗=x∗ 3 =x 6 ∗ = 0 ,x∗ 4 = 201 , x 5 ∗= 001


4.17optimum solution remains same,fn∗ew= − 27 , 600 / 3


4.19(x 1 , x 2 , x 3 , x 4 ) =number of units of products (A,B,C,D) produced.
x 1 ∗= 0004 / 3 , x 2 ∗=x 3 ∗ = 0 , x∗ 4 = 002 / 3


4.23x 1 ∗ 000 = 1 / 3 , x∗ 2 =x 3 ∗ = 0 , x∗ 4 = 008 / 3


4.29x 1 ∗ = 0 ,x 2 ∗= 0. 5 4 .31x∗ 1 = 0 , x 2 ∗= 0. 5


4 .33infinite solutions 4.35x∗ 1 = 0 ,x 2 ∗= 0. 5


4 .37 X(^2 )= { 0. 3367 , 0. 3112 , 0. 3250 }


4.40x 1 ∗= 0. 9815 , x∗ 2 = 1. 2323 , x 3 ∗= 0. 4471


CHAPTER 5


5.20.484 5.30.481 5.40.49 5.60.8 5.90.7817


5.11(a)0.786151 (b)0.786142 (c)0.786192 5.14(a) 999

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