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MATHS ( SIMPLEX METHOD)
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I answered it before... with poor typesetting xD Redo it. By the given equations: -6x - 5y - 4z + p = 0 2x + y + z + s1 = 180 x + 3y + 2z + s2 = 300 2x + y + 2z + s3 = 240 i.e. 「 ｘｙｚ s1 s2 s3ｐ| ﹁ s1 |2 1 1 1 0 0 0 | 180 | s2 |1 3 2 0 1 0 0 | 300 | s3 |2 1 2 0 0 1 0 | 240 | |－－－－－－－－－－－－－－－－－－－| p|-6 -5 -4 0 0 0 1 | 0 | ﹂」 [The large blankets are too difficult to type since i've no rights to use html script, so ignore it and imagine there are blankets.] Last row: most -ve no. = -6 => 1st column = pivot column. => R1: 180/2 = 90, R2: 300/1 = 300, R3: 240/2 = 120, that 90 is smallest. =>A11 = 2 be the pivot, and make 2 to 1. i.e. (1/2)R1 -> R1 ｘｙｚ s1 s2 s3ｐ| s1 11/21/21/20 0 0 | 90 s2 1 3 2 0 1 0 0 | 300 s3 2 1 2 0 0 1 0 | 240 －－－－－－－－－－－－－－－－－－－－ p-6 -5 -4 0 0 0 1 | 0 Make other elements in pivot column to 0. i.e. -R1 + R2 -> R2 -2R1 + R3 -> R3 6R1 + R4 -> R4 and replace the row labeling 's1' by 'x'. ｘｙｚ s1 s2 s3ｐ| x 11/21/21/2 0 0 0 | 90 s205/23/2-1/2 10 0 | 210 s3 0 0 1 -1 0 1 0 | 60 －－－－－－－－－－－－－－－－－－－－ p0 -2 -1 3 0 0 1 | 540 (previously I did these 2 step in 1 step, so it maybe quite messy xD) New pivot column = 2nd column -> 90/(1/2) = 180, 210/(5/2) = 84, 60/0 = undefined, So A22 = 5/2 be the pivot i.e. (2/5)R2 -> R2 ｘｙｚ s1 s2 s3ｐ| x 11/21/21/2 0 0 0 | 90 s2013/5-1/5 2/50 0 | 84 s3 0 0 1 -1 0 1 0 | 60 －－－－－－－－－－－－－－－－－－－－ p0 -2 -1 3 0 0 1 | 540 Again, turn others into 0: (-1/2)R2 + R1 -> R1 2R2 + R4 -> R4 Replace 's2' by 'y' ｘｙｚ s1 s2 s3ｐ| x 101/53/5 -1/5 0 0 | 48 y 013/5-1/5 2/50 0 | 84 s3 0 0 1 -1 0 1 0 | 60 －－－－－－－－－－－－－－－－－－－－ p0 01/513/54/501 | 708 So, p = 708 is maximized when x = 48, y = 84, z = 0 (and s1 = 0, s2 = 0, s3 = 60)
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MATHS ( SIMPLEX METHOD)
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maximize profit p = 6x + 5y + 4z subject to 2x + y + z ≤ 180 x + 3y + 2z ≤ 300 2x + y + 2z ≤ 240 where x, y, z are non-negative. Use simplex method to solve the question最佳解答:
I answered it before... with poor typesetting xD Redo it. By the given equations: -6x - 5y - 4z + p = 0 2x + y + z + s1 = 180 x + 3y + 2z + s2 = 300 2x + y + 2z + s3 = 240 i.e. 「 ｘｙｚ s1 s2 s3ｐ| ﹁ s1 |2 1 1 1 0 0 0 | 180 | s2 |1 3 2 0 1 0 0 | 300 | s3 |2 1 2 0 0 1 0 | 240 | |－－－－－－－－－－－－－－－－－－－| p|-6 -5 -4 0 0 0 1 | 0 | ﹂」 [The large blankets are too difficult to type since i've no rights to use html script, so ignore it and imagine there are blankets.] Last row: most -ve no. = -6 => 1st column = pivot column. => R1: 180/2 = 90, R2: 300/1 = 300, R3: 240/2 = 120, that 90 is smallest. =>A11 = 2 be the pivot, and make 2 to 1. i.e. (1/2)R1 -> R1 ｘｙｚ s1 s2 s3ｐ| s1 11/21/21/20 0 0 | 90 s2 1 3 2 0 1 0 0 | 300 s3 2 1 2 0 0 1 0 | 240 －－－－－－－－－－－－－－－－－－－－ p-6 -5 -4 0 0 0 1 | 0 Make other elements in pivot column to 0. i.e. -R1 + R2 -> R2 -2R1 + R3 -> R3 6R1 + R4 -> R4 and replace the row labeling 's1' by 'x'. ｘｙｚ s1 s2 s3ｐ| x 11/21/21/2 0 0 0 | 90 s205/23/2-1/2 10 0 | 210 s3 0 0 1 -1 0 1 0 | 60 －－－－－－－－－－－－－－－－－－－－ p0 -2 -1 3 0 0 1 | 540 (previously I did these 2 step in 1 step, so it maybe quite messy xD) New pivot column = 2nd column -> 90/(1/2) = 180, 210/(5/2) = 84, 60/0 = undefined, So A22 = 5/2 be the pivot i.e. (2/5)R2 -> R2 ｘｙｚ s1 s2 s3ｐ| x 11/21/21/2 0 0 0 | 90 s2013/5-1/5 2/50 0 | 84 s3 0 0 1 -1 0 1 0 | 60 －－－－－－－－－－－－－－－－－－－－ p0 -2 -1 3 0 0 1 | 540 Again, turn others into 0: (-1/2)R2 + R1 -> R1 2R2 + R4 -> R4 Replace 's2' by 'y' ｘｙｚ s1 s2 s3ｐ| x 101/53/5 -1/5 0 0 | 48 y 013/5-1/5 2/50 0 | 84 s3 0 0 1 -1 0 1 0 | 60 －－－－－－－－－－－－－－－－－－－－ p0 01/513/54/501 | 708 So, p = 708 is maximized when x = 48, y = 84, z = 0 (and s1 = 0, s2 = 0, s3 = 60)
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