1. **Problem statement:** A farmer has 24 meters of fencing to enclose a garden along the walls of an L-shaped house. The garden is divided into three equal sections for carrots, tomatoes, and cucumbers. We need to find the maximum total area of the garden and the dimensions that maximize this area. 2. **Understanding the setup:** Since the garden is against two walls forming an L-shape, fencing is only needed for the outer sides and the internal dividers between the three sections. 3. **Define variables:** Let $x$ be the length of the side perpendicular to the house walls, and $y$ be the length parallel to the walls. The garden is divided into three equal sections by two fences parallel to $x$, so there are 4 vertical fence segments of length $x$ and 1 horizontal fence segment of length $y$. 4. **Fence length equation:** Total fence used is the sum of the 4 vertical fences and 1 horizontal fence: $$4x + y = 24$$ 5. **Area equation:** The total area $A$ of the garden is: $$A = x \times y$$ 6. **Express $y$ in terms of $x$:** From the fence equation, $$y = 24 - 4x$$ 7. **Area as a function of $x$:** Substitute $y$ into the area formula: $$A(x) = x(24 - 4x) = 24x - 4x^2$$ 8. **Maximize the area:** Take the derivative of $A(x)$ with respect to $x$ and set it to zero: $$\frac{dA}{dx} = 24 - 8x = 0$$ 9. **Solve for $x$:** $$8x = 24 \implies x = 3$$ 10. **Find $y$:** $$y = 24 - 4(3) = 24 - 12 = 12$$ 11. **Calculate maximum area:** $$A = 3 \times 12 = 36$$ 12. **Conclusion:** The maximum area of the garden is 36 square meters when the side perpendicular to the walls is 3 meters and the side parallel to the walls is 12 meters.