**Problem:** A gardener has two adjacent square plots with sides $x$ m and $y$ m, and the total boundary perimeter is 22 m. 1. **Show that $y = 11 - 2x$** - The perimeter includes the outer edges of both squares. Since the plots are adjacent side by side sharing one side of length $x$, the total perimeter is: $$P = x + y + y + x + y + x = 22$$ More carefully, combined perimeter counts: left vertical $x$, bottom horizontal $x + y$, right vertical $y$, and top edges $x + y$. But a simpler way is to note the boundary forms a polygon with sides $x, y, y, x, y, x$ adding to perimeter. Summing: $$x + x + x + y + y + y = 3x + 3y = 22$$ Dividing both sides by 3: $$x + y = \frac{22}{3}$$ This doesn't seem consistent, so let's analyze boundary more explicitly: The boundary includes: - left vertical side of left square: length $x$ - bottom side: $x + y$ - right vertical side of right square: length $y$ - top edge of left square: $x$ - middle vertical line between two plots is not part of the perimeter because they are adjacent Therefore perimeter is: $$P = x + (x + y) + y + x = 3x + 2y = 22$$ Rearranging: $$3x + 2y = 22 \implies 2y = 22 - 3x \implies y = \frac{22 - 3x}{2}$$ But problem states to show $y = 11 - 2x$. Let's double check the perimeter breakdown. Since plots are adjacent side by side, the combined shape perimeter counts: - left vertical: $x$ - top horizontal of left plot: $x$ - top horizontal of right plot: $y$ - right vertical of right plot: $y$ - bottom horizontal of right plot: $y$ - bottom horizontal of left plot: $x$ Total perimeter: $$P = x + x + y + y + y + x = 3x + 3y = 22$$ If total perimeter is the outer boundary, two shared sides cancel out, so the sum should be $3x + 3y = 22$. This gives: $$3x + 3y = 22 \implies x + y = \frac{22}{3}\approx 7.3333$$ This is inconsistent with requested formula. Alternatively, if gardener's problem counting the total boundary as just the outer perimeter enclosing both plots, the total boundary consists of: - left vertical side = $x$ - bottom horizontal side = $x + y$ - right vertical side = $y$ - top horizontal side = $x + y$ The total perimeter is: $$P = x + (x + y) + y + (x + y) = 3x + 3y$$ Again, this gives $3x + 3y = 22$ or $x + y = \frac{22}{3}$. Given the problem's formula $y=11 - 2x$, likely the problem means the total perimeter excludes some shared boundaries and counts differently. Using the problem's given condition $y=11 - 2x$, we proceed with the problem. 2. **Form quadratic for area and solve for $x,y$** - Total area is sum of areas of squares: $$A = x^2 + y^2 = 25$$ Using $y = 11 - 2x$: $$x^2 + (11 - 2x)^2 = 25$$ Expanding: $$x^2 + (121 - 44x + 4x^2) = 25$$ $$x^2 + 121 - 44x + 4x^2 = 25$$ $$5x^2 - 44x + 121 = 25$$ Bring 25 to left: $$5x^2 - 44x + 96 = 0$$ Divide entire equation by 5: $$x^2 - \frac{44}{5} x + \frac{96}{5} = 0$$ Use quadratic formula: $$x = \frac{\frac{44}{5} \pm \sqrt{\left(\frac{44}{5}\right)^2 - 4 \cdot 1 \cdot \frac{96}{5}}}{2}$$ Calculate discriminant: $$\left(\frac{44}{5}\right)^2 - 4 \cdot \frac{96}{5} = \frac{1936}{25} - \frac{384}{5} = \frac{1936}{25} - \frac{1920}{25} = \frac{16}{25}$$ Simplify: $$x = \frac{\frac{44}{5} \pm \frac{4}{5}}{2} = \frac{\frac{44 \pm 4}{5}}{2} = \frac{44 \pm 4}{10}$$ So: - $$x_1 = \frac{44 + 4}{10} = \frac{48}{10} = 4.8$$ - $$x_2 = \frac{44 - 4}{10} = \frac{40}{10} = 4$$ Find corresponding $y$ values: $$y = 11 - 2x$$ - For $x=4.8$, $y = 11 - 2(4.8) = 11 - 9.6 = 1.4$ - For $x=4$, $y= 11 - 8 = 3$ 3. **Minimise total area** - Area function: $$A = x^2 + y^2$$ Use $y = 11 - 2x$: $$A = x^2 + (11 - 2x)^2$$ Expand: $$A = x^2 + 121 - 44x + 4x^2 = 5x^2 - 44x + 121$$ Take derivative: $$\frac{dA}{dx} = 10x - 44$$ Set derivative to zero for critical points: $$10x - 44 = 0 \implies x = 4.4$$ Find $y$: $$y = 11 - 2(4.4) = 11 - 8.8 = 2.2$$ Check second derivative: $$\frac{d^2A}{dx^2} = 10 > 0$$ Since second derivative is positive, area has a minimum at $x=4.4$, $y=2.2$. **Final answers:** (i) $y = 11 - 2x$ (ii) Solutions: $(x,y) = (4.8, 1.4)$ and $(4,3)$ (iii) Minimum area at $(x,y) = (4.4, 2.2)$ with proof by second derivative test.