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Unformatted text preview: CE 335 Solutions to Homework 5 11) We need to assume that all material supplied is used. Augmented matrix formulation of problem: 15 17 19  3890 1.0 1.2 1.5  282 0.3 0.4 0.55  95 subtract multiple of first row from second and third (in Matlab: A(2, :) = A(2, :)  (A(2, 1 )/A(1, 1))*A(1, :); A(3, :) = A(3, :)  (A(3, 1)/A(1, 1))*A(1, :) ): 15 17 19  3890 1/15 7/30  68/3 0.06 0.17  17.2 subtract a multiple of the second row from the third (in Matlab: A(3, :) = A(3, :)  (A(3, 2 )/A(2, 2))*A(2, :) ): 15 17 19  3890 1/15 7/30  68/30.04 3.2 with backsubstitution, we now get x 3 = 80, x 2 = 60, x 1 = 90 for the number of each component that can be produced per day. 12) We use the finitedifference approximations c''(x) (c(x + h)  2c(x) + c(x  h)) / h 2 c'(x) (c(x + h)  c(x  h)) / 2h so from the differential equation (D/h 2 )(c(x + h)  2c(x) + c(x  h))  (U/2h)(c(x + h)  c(x  h))  kc(x) 0 Taking, for example, h = 0.5, we consider 21 values of c between x = 0 and x = 10, where c(0) and c(10Taking, for example, h = 0....
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 Spring '11
 Lybas

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