1. **Problem:** A rectangular garden is to be constructed using a rock wall as one side and wire fencing for the other three sides. Given 200 m of wire fence, find the dimensions that maximize the garden's area and the maximum area. 2. **Formula and rules:** - Let the length parallel to the rock wall be $L$ and the width perpendicular be $W$. - The fencing is used for three sides: two widths and one length, so the fencing constraint is: $$2W + L = 200$$ - The area $A$ of the rectangle is: $$A = L \times W$$ 3. **Express $L$ in terms of $W$ using the fencing constraint:** $$L = 200 - 2W$$ 4. **Substitute $L$ into the area formula:** $$A = (200 - 2W)W = 200W - 2W^2$$ 5. **Maximize $A$ by finding critical points:** - Take derivative with respect to $W$: $$\frac{dA}{dW} = 200 - 4W$$ - Set derivative to zero: $$200 - 4W = 0$$ $$4W = 200$$ $$W = 50$$ 6. **Find corresponding $L$:** $$L = 200 - 2(50) = 200 - 100 = 100$$ 7. **Check that this is a maximum:** - Second derivative: $$\frac{d^2A}{dW^2} = -4 < 0$$ - Negative second derivative confirms a maximum. 8. **Calculate maximum area:** $$A_{max} = L \times W = 100 \times 50 = 5000$$ **Final answer:** - Dimensions: Length $L = 100$ m, Width $W = 50$ m - Maximum area: $5000$ m$^2$