A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 1900 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions? The length of the shorter side of the rectangular plot is The length of the longer side of the rectangular plot is

To maximize the area of a rectangular plot of farmland bounded on one side by a river and enclosed by a fence on the other three sides, we can set up the problem as follows: Let: - \( L \) be the length of the side parallel to the river (the longer side). - \( W \) be the width of the plot (the shorter side). Since the river bounds one side, we only need to use the wire for the other three sides. The total length of the wire used for the fence is given by: \[ L + 2W = 1900 \] We want to maximize the area \( A \) of the rectangle, which is given by: \[ A = L \times W \] From the wire constraint, we can express \( L \) in terms of \( W \): \[ L = 1900 - 2W \] Now, substitute this expression for \( L \) into the area formula: \[ A = (1900 - 2W) \times W \] Expanding this gives: \[ A = 1900W - 2W^2 \] This is a quadratic equation in the standard form \( A = -2W^2 + 1900W \). To find the maximum area, we can use the vertex formula for a parabola, which occurs at: \[ W = -\frac{b}{2a} \] where \( a = -2 \) and \( b = 1900 \): \[ W = -\frac{1900}{2 \times -2} = \frac{1900}{4} = 475 \] Now, substituting \( W = 475 \) back into the equation for \( L \): \[ L = 1900 - 2(475) = 1900 - 950 = 950 \] Thus, the dimensions of the rectangular plot that maximize the area are: - The length of the shorter side (width) is \( W = 475 \) m. - The length of the longer side (length) is \( L = 950 \) m. In summary: - The length of the shorter side of the rectangular plot is **475 m**. - The length of the longer side of the rectangular plot is **950 m**.