# ch18 - CHAPTER 18 SOLUTIONS 18.1 The unknowns are A and B,...

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18-1 CHAPTER 18 SOLUTIONS 18.1 The unknowns are A and B, the quantities of A and B produced in gal/week. The objective function is: A,B (1) max 5A + 3.5B J = The constraints are: AB (2) 2 6,000 [Type 1 columns are available 6,000 hrs/week] 100 100 i.e. 2A + B 600,000 (3) 4 10,000 [Type 2 columns are available 10,000 hrs/week] 100 100 i.e. A + 4B 1,000,000 + + The graphical solution of the LP is shown above, were it is noted that the feasible region is constrained by the quantities of the two types of columns that are available [Eqs. (2) and (3)]. The three coordinates of the feasible region boundaries are possible optimal solutions. Their examination gives: 1: A = 0 gal, B = 250,000 gal J = \$875,00/week 2: A = 200,000 gal B = 200,000 gal J = \$1,700,000/week 3: A = 300,000 gal B = 0 gal J = \$1,500,000/week Clearly, the optimal solution is to produce 200,000 gal each of A and B, for which the profit is maximized, at \$1,700,000/week.

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18-2 18.2 (a) The function is: { } 32 2 12 1 1 2 1 2 2 ,2 41 0 fxx x xx xx x =+ + = +− (1) Differential analysis of the function gives: 22 1 2 1 2 2 2 11 2 21 2 2 641 0 0 81 0 2 12 8 10 0 82 f x x f f x x x ff x H x   ∇= = = −+  ∂∂ == The equations for 0 f are both of second order, indicating four possible solutions. The second equation gives: ( ) 2 0 8 x =− (2) Substituting into the first equation gives a fourth order polynomial in x 2 : 43 2 256 1,280 2,024 1,000 0 −= = (3) The roots of Eq. (3) are obtained using MATLAB, and substituted into Eq. (2) to obtain values of x 1 , and into Eq. (1), to obtain values of the function. The results are summarized in the table. Contour lines for the function are plotted below. x 1 x 2 f Nature of Extremum 0.0000 0.0000 0.0000 Saddle ( H has one positive and one negative eigenvalue) 1.0000 1.0000 -3.0000 Minimum ( H has two positive eigenvalues) -0.9541 1.6938 6.3434 Maximum ( H has two negative eigenvalues) -0.5459 2.3062 5.9691 Saddle ( H has one positive and one negative eigenvalue)
18-3 (b) The function is: { } 911 5 12 1 2 , 1,000 4 10 2.5 10 fxx x xx x −− =+ × + × ×+ (1) Differential analysis of the function gives: 921 1 912 5 2 22 2 31 9 11 2 13 2 2 1,000 4 10 0 41 0 2 . 51 0 2 0 2 f x f f x ff x H x x x   −× ∇= = = = = + × ×  ∂∂ == × The first equation of 0 f gives: 62 21 0 x x = × (2) Substituting into the second equation gives a polynomial in x 1 : 39 2 10 1,000. From Eq. (1), 4 in both cases. x = = ± = ⇒= ± = (3) The solutions to 0 f are substituted in Eq. (1) and are summarized in the table below. x 1 x 2 f Nature of Extremum 1,000 4 -10 6 M a x i m u m ( H has two negative eigenvalues) -1,000 4 3 × 10 6 Minimum ( H has two positive eigenvalues)

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18-4 (c) The function is: { }() ( ) 2 2 2 12 1 2 1 , 1 100 fxx x x x =− + (1) This is Rosenbrock’s famous, banana function.” Differential analysis of the function gives: ( ) ( ) ( ) ( ) 2 11 2 1 1 2 21 2 22 2 2 2 1 1 2 2 2 1 400 0 200 2 400 3 400 400 200 f xx x x x f f x ff x x H x x   −−− ∇= = =  ∂∂ ∂ −− −− == The second equation of 0 f gives: 2 x x = (2) Substituting into the second equation gives a polynomial in x 1 : ( ) ( ) 1 1 1 2 2 1 400 1. From Eq. (1), 1. x x x x = = ⇒= = (3) The solutions to 0 f are substituted in Eq. (1) and are summarized in the table
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## This note was uploaded on 04/10/2008 for the course CHE 233W taught by Professor Debelak during the Spring '08 term at Vanderbilt.

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ch18 - CHAPTER 18 SOLUTIONS 18.1 The unknowns are A and B,...

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