1. **Problem statement:** (a) Given a rectangular garden with area 80 m², width $y$ metres, length $x$ metres, and a path of uniform width 2 metres around it. (b) Find an expression for $y$ in terms of $x$. (c) Show the total area of the garden plus path is $A = 96 + 4x + \frac{320}{x}$. (d) Find the minimum area of the path. 2. **Step (a): Express $y$ in terms of $x$** - The area of the garden is given by $xy = 80$. - Solve for $y$: $$y = \frac{80}{x}$$ 3. **Step (b): Expression for total area $A$ including the path** - The path adds 2 metres on each side, so the total length including the path is $x + 2 \times 2 = x + 4$. - Similarly, the total width including the path is $y + 4$. - Total area $A$ is: $$A = (x + 4)(y + 4)$$ - Substitute $y = \frac{80}{x}$: $$A = (x + 4)\left(\frac{80}{x} + 4\right)$$ - Expand the expression: $$A = (x + 4)\left(\frac{80 + 4x}{x}\right) = \frac{(x + 4)(80 + 4x)}{x}$$ - Multiply numerator: $$(x + 4)(80 + 4x) = 80x + 4x^2 + 320 + 16x = 4x^2 + 96x + 320$$ - So, $$A = \frac{4x^2 + 96x + 320}{x} = 4x + 96 + \frac{320}{x}$$ - Rearranged as: $$A = 96 + 4x + \frac{320}{x}$$ 4. **Step (c): Find the minimum area of the path** - The area of the path alone is total area minus garden area: $$A_{path} = A - 80 = 96 + 4x + \frac{320}{x} - 80 = 16 + 4x + \frac{320}{x}$$ - To minimize $A_{path}$, differentiate with respect to $x$ and set to zero: $$\frac{dA_{path}}{dx} = 4 - \frac{320}{x^2} = 0$$ - Solve for $x$: $$4 = \frac{320}{x^2} \implies 4x^2 = 320 \implies x^2 = 80 \implies x = \sqrt{80} = 4\sqrt{5}$$ - Substitute $x$ back into $A_{path}$: $$A_{path} = 16 + 4(4\sqrt{5}) + \frac{320}{4\sqrt{5}} = 16 + 16\sqrt{5} + \frac{320}{4\sqrt{5}}$$ - Simplify the last term: $$\frac{320}{4\sqrt{5}} = \frac{80}{\sqrt{5}} = 80 \times \frac{\sqrt{5}}{5} = 16\sqrt{5}$$ - So, $$A_{path} = 16 + 16\sqrt{5} + 16\sqrt{5} = 16 + 32\sqrt{5}$$ - Numerically, $\sqrt{5} \approx 2.236$, so: $$A_{path} \approx 16 + 32 \times 2.236 = 16 + 71.55 = 87.55$$ - Therefore, the minimum area of the path is approximately 87.55 m². **Final answers:** - (a) $y = \frac{80}{x}$ - (b) $A = 96 + 4x + \frac{320}{x}$ - (c) Minimum path area $\approx 87.55$ m² at $x = 4\sqrt{5}$