The feasible region determined by some constraints is represented by the shaded region in the graph…
The feasible region determined by some constraints is represented by the shaded region in the graph given below:

(i)[1.0]
Formulate the constraints which represent the above feasible region.
(ii)[1.0]
Hence, maximise the objective function given by $Z = x + y$.
(iii)[1.0]
What change in the constraints will make the feasible region unbounded?
Answer
Answer (i)
AIWritten by AI - it can contain mistakes.
The boundary lines of the feasible region are:
1. Line through $A(1,1)$ and $B(3,3)$: equation is $y = x \Rightarrow x - y \le 0$ (or $y - x \ge 0$).
2. Line through $B(3,3)$ and $C(1,7)$: slope is $\frac{7-3}{1-3} = -2$, equation is $y - 3 = -2(x - 3) \Rightarrow 2x + y \le 9$.
3. Vertical line through $A(1,1)$ and $C(1,7)$: $x \ge 1$.
4. Non-negativity constraints: $x \ge 0, y \ge 0$.
Therefore, the constraints representing the feasible region are:
$2x + y \le 9$, $x - y \le 0$, $x \ge 1$, $x \ge 0, y \ge 0$.
Answer (ii)
AIWritten by AI - it can contain mistakes.
The corner points of the bounded feasible region are $A(1,1)$, $B(3,3)$, and $C(1,7)$.
Evaluating the objective function $Z = x + y$ at each vertex:
- At $A(1,1)$: $Z = 1 + 1 = 2$
- At $B(3,3)$: $Z = 3 + 3 = 6$
- At $C(1,7)$: $Z = 1 + 7 = 8$
Therefore, the maximum value of $Z$ is $8$, which occurs at $C(1,7)$.
Final answer: 8
Answer (iii)
AIWritten by AI - it can contain mistakes.
Changing the constraint $2x + y \le 9$ to $2x + y \ge 9$ (or removing the upper bound constraint $2x + y \le 9$) will make the feasible region unbounded.
From ISC 2027 Specimen Mathematics Paper 1, question 9.