# Question 4:

## Part (a):
| $5x + 3y + 10z \leq 15$ (or equivalent from row 2) | B1 | One correct inequality from tableau rows |
| $7x + 6y + 4z \leq 28$ (or equivalent from row 3) | B1 | Second and third correct inequalities |
| $4x + 3y + 6z \leq 12$ (or equivalent from row 4) | | |

## Part (b)(i):
| Pivot column is $y$, ratios: $15/3=5$, $28/6=4\frac{2}{3}$, $12/3=4$, so pivot row is row 4, pivot element is 3 | M1 | Correct identification of pivot |
| New row 4: $0, \frac{4}{3}, 1, 2, 0, 0, \frac{1}{3}, 4$ | A1 | Correct pivot row after division |
| New objective row: $1, 6, 0, k-12, 0, 0, 2, 24$ | A1 | Correct objective row |
| Remaining rows updated correctly | A1 | All rows correct |

Updated tableau:

| $P$ | $x$ | $y$ | $z$ | $s$ | $t$ | $u$ | value |
|---|---|---|---|---|---|---|---|
| $1$ | $6$ | $0$ | $k-12$ | $0$ | $0$ | $2$ | $24$ |
| $0$ | $3$ | $0$ | $4$ | $1$ | $0$ | $-1$ | $3$ |
| $0$ | $3$ | $0$ | $-8$ | $0$ | $1$ | $-2$ | $4$ |
| $0$ | $\frac{4}{3}$ | $1$ | $2$ | $0$ | $0$ | $\frac{1}{3}$ | $4$ |

## Part (b)(ii):
| For optimal not reached, need at least one negative value in objective row | M1 | Correct reasoning |
| $k - 12 < 0$ or $6 < 0$ (impossible), so $k < 12$ | A1 | $k < 12$ |

## Part (c)(i): (when $k=20$)
| Objective row becomes $1, 6, 0, 8, 0, 0, 2, 24$; pivot column is $x$, ratios: $3/3=1$, $4/3$, $4/(\frac{4}{3})=3$; pivot element is 3 in row 1 | M1 | Correct pivot identified |
| New pivot row: $0, 1, 0, \frac{4}{3}, \frac{1}{3}, 0, -\frac{1}{3}, 1$ | A1 | Correct row |
| New objective row: $1, 0, 0, \ldots, -2, 0, 4, 18$ | A1 | |
| All rows correct | A1 | Fully correct tableau |

## Part (c)(ii):
| $P = 18$, $x=1$, $y=4$, $z=0$ | B1 | Correct values stated |
| Slack variables: $s=0$, $t=0$, $u=0$ | B1 | All slack variables correct |
| Constraints 1, 2 and 3 are all binding (active) | B1 | Correct interpretation |

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