A group needs to model the constraints and cost for a chalet booking. The constraints are: at least 4 Type A, at least 4 Type B, the ratio of Type A to Type B is at least 1:2, and the total number of

The gradient of the search line is $\boxed{-\frac{2}{3}}$.

1.2.4

a) Using a scale of 1 unit = 2 chalets, represent the inequalities graphically on the attached ADDENDUM A (1.2.4).
The inequalities to be graphed are:

(A graph is required for this part, which is not provided. The lines to be drawn are $x=4$, $y=4$, $y=2x$, and $x+y=18$.)

b) Indicate the feasible region on the graph.
(The feasible region is the area on the graph that satisfies all the inequalities simultaneously. It is a polygon formed by the intersection of the constraint lines.)

c) Draw the search line on the graph and calculate the maximum cost for their stay.

The constraint lines are $x=4$, $y=4$, $y=2x$, and $x+y=18$.

• Intersection of $x=4$ and $y=4$: $(4,4)$
• Intersection of $x=4$ and $y=2x$: $(4, 2(4)) = (4,8)$
• Intersection of $y=4$ and $x+y=18$: $x+4=18 \implies x=14$. So $(14,4)$
• Intersection of $y=2x$ and $x+y=18$: Substitute $y=2x$ into $x+y=18 \implies x+2x=18 \implies 3x=18 \implies x=6$. Then $y=2(6)=12$. So $(6,12)$ The vertices of the feasible region are $(4,4)$, $(4,8)$, $(6,12)$, and $(14,4)$.

• At $(4,4)$: $C = 1000(4) + 1500(4) = 4000 + 6000 = 10000$
• At $(4,8)$: $C = 1000(4) + 1500(8) = 4000 + 12000 = 16000$
• At $(6,12)$: $C = 1000(6) + 1500(12) = 6000 + 18000 = 24000$
• At $(14,4)$: $C = 1000(14) + 1500(4) = 14000 + 6000 = 20000$

The maximum cost is the largest value obtained.
The maximum cost for their stay is $\boxed{\text{R24 000}}$.

1.2.5 Due to cancellations of tourists, the group on arrival consisted of 10 families who needed type B chalets and 3 couples who required type A chalets. Will the tour leader be able to meet the requirements of all the tourist in their booked chalets? Give a reason for your answer with reference to the feasible region.

The new requirements are $x=3$ (3 couples for Type A) and $y=10$ (10 families for Type B). This corresponds to the point $(3,10)$.

The constraints are:

1. $x \ge 4$: For $(3,10)$, $3 \ge 4$ is False.
2. $y \ge 4$: For $(3,10)$, $