1. **State the problem:** We need to find the values of $x$ and $y$ that minimize the area of an L/T-shaped fence with a fixed perimeter of 300 meters. 2. **Identify variables and expressions:** The shape consists of segments with lengths $x$ and $y$ as labeled. 3. **Write the perimeter equation:** The total perimeter is given as 300 meters. The perimeter includes the outer edges of the shape. From the diagram description, the perimeter is composed of: - Two vertical sides of length $x$ - Two vertical sides of length $y$ - One horizontal base of length $4y$ - Two horizontal sides of length $x$ Summing these, the perimeter $P$ is: $$P = 2x + 2y + 4y + 2x = 4x + 6y$$ Given $P = 300$, we have: $$4x + 6y = 300$$ 4. **Express one variable in terms of the other:** Solve for $x$: $$4x = 300 - 6y$$ $$x = \frac{300 - 6y}{4} = 75 - \frac{3}{2}y$$ 5. **Write the area expression:** The area $A$ consists of the sum of the rectangular parts: - The top rectangle: $x \times x = x^2$ - The bottom rectangle: $4y \times y = 4y^2$ So, $$A = x^2 + 4y^2$$ 6. **Substitute $x$ in terms of $y$ into the area:** $$A = \left(75 - \frac{3}{2}y\right)^2 + 4y^2$$ Expand the square: $$A = \left(75\right)^2 - 2 \times 75 \times \frac{3}{2}y + \left(\frac{3}{2}y\right)^2 + 4y^2$$ $$A = 5625 - 225y + \frac{9}{4}y^2 + 4y^2$$ Combine like terms: $$A = 5625 - 225y + \left(\frac{9}{4} + 4\right) y^2 = 5625 - 225y + \frac{25}{4} y^2$$ 7. **Minimize the area:** Take the derivative of $A$ with respect to $y$ and set it to zero: $$\frac{dA}{dy} = -225 + \frac{25}{2} y = 0$$ Solve for $y$: $$\frac{25}{2} y = 225$$ $$y = \frac{225 \times 2}{25} = 18$$ 8. **Find $x$ using $y=18$:** $$x = 75 - \frac{3}{2} \times 18 = 75 - 27 = 48$$ 9. **Verify the perimeter:** $$4x + 6y = 4 \times 48 + 6 \times 18 = 192 + 108 = 300$$ 10. **Final answer:** $$x = 48, \quad y = 18$$ These values minimize the area with the given perimeter constraint.