1. **Problem statement:** A gardener has 18 metres of timber fencing to enclose a rectangular vegetable patch using a straight stone wall as one side. The width is $x$ metres. We need to show the area $A(x)$ of the enclosure is given by $A(x) = 18x - 2x^2$. 2. **Deriving the area function:** - Let the width be $x$ metres. - Since one side is a stone wall, fencing is needed for the other three sides: two widths and one length. - Total fencing used: $2x + L = 18$ metres, where $L$ is the length. - Solve for $L$: $$L = 18 - 2x$$ - Area $A$ is width times length: $$A = x \times L = x(18 - 2x) = 18x - 2x^2$$ 3. **Graph of $y = A(x)$ for $0 \leq x \leq 9$:** - The domain is $0 \leq x \leq 9$ because $x$ cannot be negative and $L$ must be non-negative. 4. **(i) Find area when $x=3$:** $$A(3) = 18(3) - 2(3)^2 = 54 - 18 = 36$$ 5. **(ii) Find $x$ when area is 30:** - Set $A(x) = 30$: $$18x - 2x^2 = 30$$ - Rearrange: $$-2x^2 + 18x - 30 = 0$$ - Divide both sides by $-2$: $$\cancel{-2}x^2 - \cancel{18}x + \cancel{30} = 0 \Rightarrow x^2 - 9x + 15 = 0$$ - Use quadratic formula: $$x = \frac{9 \pm \sqrt{9^2 - 4 \times 1 \times 15}}{2} = \frac{9 \pm \sqrt{81 - 60}}{2} = \frac{9 \pm \sqrt{21}}{2}$$ - Approximate: $$x \approx \frac{9 \pm 4.58}{2}$$ - So, $$x_1 \approx \frac{9 - 4.58}{2} = 2.21, \quad x_2 \approx \frac{9 + 4.58}{2} = 6.79$$ 6. **(iii) Find length and breadth for maximum area:** - $A(x) = 18x - 2x^2$ is a quadratic with a negative leading coefficient, so maximum at vertex. - Vertex $x$ coordinate: $$x = -\frac{b}{2a} = -\frac{18}{2 \times (-2)} = \frac{18}{4} = 4.5$$ - Length: $$L = 18 - 2(4.5) = 18 - 9 = 9$$ - Breadth is $x = 4.5$ 7. **(iv) Maximum area:** $$A(4.5) = 18(4.5) - 2(4.5)^2 = 81 - 40.5 = 40.5$$ **Final answers:** - Area at $x=3$ is 36 m². - Values of $x$ for area 30 m² are approximately 2.21 m and 6.79 m. - Maximum area is 40.5 m² with length 9 m and breadth 4.5 m.