The Garden Gate
A rectangular garden is to be enclosed by a fence on three sides, with the fourth side being a wall. The fence material costs 10 dollars per meter, and the wall is already there. The garden must have an area of 100 square meters. The gardener wants to minimize the cost of the fence. However, there is a constraint: the length of the garden (parallel to the wall) must be at least 5 meters and at most 20 meters. The gardener also considers adding a gate that costs a fixed 50 dollars, but the gate does not affect the fencing dimensions.
1. Determine the dimensions of the garden that minimize the fencing cost, ignoring the gate. What is the minimum cost?
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Let the length parallel to the wall be (x) meters, and the width perpendicular to the wall be (y) meters. The area constraint is (xy = 100). The fence is needed for two widths and one length, so the total length of fence is (L = x + 2y). The cost is (C = 10L = 10(x + 2y)). From (xy = 100), we have (y = 100/x). Then (C(x) = 10\left(x + \frac{200}{x}\right)). To minimize, take derivative: (C'(x) = 10\left(1 - \frac{200}{x^2}\right)). Set to zero: (1 = \frac{200}{x^2}), so (x^2 = 200), (x = \sqrt{200} = 10\sqrt{2} \approx 14.14) m. Then (y = 100/(10\sqrt{2}) = 10/\sqrt{2} = 5\sqrt{2} \approx 7.07) m. Check constraints: (5 \le x \le 20), and (10\sqrt{2} \approx 14.14) is within [5,20], so it's valid. The minimum cost is (C = 10(10\sqrt{2} + 2 \cdot 5\sqrt{2}) = 10(10\sqrt{2} + 10\sqrt{2}) = 10 \cdot 20\sqrt{2} = 200\sqrt{2} \approx 282.84) dollars. Since the second derivative (C''(x) = 10 \cdot \frac{400}{x^3} > 0) for (x>0), this is a minimum.
2. If the gardener decides to include the gate, what is the new minimum total cost? Does the optimal dimension change?
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The gate adds a fixed cost of 50 dollars, so the total cost becomes (C_{\text{total}} = 10(x + 2y) + 50). Since the gate cost is constant, it does not affect the optimization of the fencing dimensions. The minimum fencing cost remains (200\sqrt{2}) dollars, so the new minimum total cost is (200\sqrt{2} + 50 \approx 282.84 + 50 = 332.84) dollars. The optimal dimensions remain (x = 10\sqrt{2}) m and (y = 5\sqrt{2}) m.