Optimization of a Rectangular Enclosure
A farmer has 100 meters of fencing to enclose a rectangular field adjacent to a straight river. No fencing is needed along the river. The farmer wants to maximize the area of the field. Let the length of the field parallel to the river be (x) meters and the width perpendicular to the river be (y) meters. The fencing is used for the two widths and one length, so (x + 2y = 100). The area is (A = xy). The farmer also considers building a second identical enclosure next to the first, sharing the side parallel to the river, but that would require additional fencing and may not be feasible with the available 100 m.
1. Determine the dimensions that maximize the area of the single rectangular field. Show your reasoning using calculus or algebra.
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From the constraint (x + 2y = 100), we have (x = 100 - 2y). The area is (A = xy = (100 - 2y)y = 100y - 2y^2). This is a quadratic function opening downward. The maximum occurs at the vertex: (y = -\frac{b}{2a} = -\frac{100}{2(-2)} = 25). Then (x = 100 - 2(25) = 50). So the dimensions are 50 m (parallel to river) by 25 m (perpendicular). The maximum area is (50 \times 25 = 1250, \text{m}^2). Alternatively, using calculus: (\frac{dA}{dy} = 100 - 4y = 0 \implies y = 25), and (\frac{d^2A}{dy^2} = -4 < 0), confirming a maximum.
2. Suppose the farmer wants to build two identical adjacent rectangular enclosures side by side, sharing the side parallel to the river. The total fencing available is still 100 m, and the shared side does not require fencing (since it's along the river? Actually, the shared side is perpendicular to the river? Clarify: The enclosures are adjacent, sharing a side perpendicular to the river. The fencing is used for the outer boundary and the shared side. Let each enclosure have length \(x\) parallel to the river and width \(y\) perpendicular. The total fencing used is \(2x + 3y = 100\). Find the dimensions that maximize the total area of the two enclosures combined.
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The total area is (A = 2xy). The constraint is (2x + 3y = 100), so (x = \frac{100 - 3y}{2}). Then (A = 2 \left(\frac{100 - 3y}{2}\right) y = (100 - 3y)y = 100y - 3y^2). This is a quadratic in (y). The maximum occurs at (y = -\frac{100}{2(-3)} = \frac{100}{6} = \frac{50}{3} \approx 16.67, \text{m}). Then (x = \frac{100 - 3(50/3)}{2} = \frac{100 - 50}{2} = 25, \text{m}). So each enclosure is 25 m by 16.67 m, and total area is (2 \times 25 \times 16.67 \approx 833.33, \text{m}^2). This is less than the single enclosure area, so the farmer should not build two enclosures if maximizing area with the same fencing.