sol_hw5 - CE 335 Solutions to Homework 5 11) We need to...

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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 | 68/3-0.04 |-3.2 with back-substitution, 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 finite-difference 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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sol_hw5 - CE 335 Solutions to Homework 5 11) We need to...

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