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 maximum area of the rectangular plot is

To find the largest area that can be enclosed by the given conditions, we need to consider the perimeter of the rectangular plot. Since the plot is bounded on one side by a river and on the other three sides by a single-strand electric fence, the perimeter of the rectangular plot will be the sum of the lengths of the sides that are bounded by the fence. Let's denote the length of the side parallel to the river as \( l \) and the length of the side perpendicular to the river as \( w \). The perimeter of the rectangular plot will be \( 2l + w \). Given that we have 1900 m of wire at our disposal, the perimeter of the rectangular plot will be equal to 1900 m. Therefore, we have the equation: \[ 2l + w = 1900 \] We want to maximize the area of the rectangular plot, which is given by the formula: \[ A = lw \] To find the maximum area, we need to express the area in terms of one variable and then find the maximum value of that variable. Let's solve for \( w \) in terms of \( l \) from the perimeter equation: \[ w = 1900 - 2l \] Now, substitute this expression for \( w \) into the area formula: \[ A = l(1900 - 2l) \] To find the maximum area, we need to find the critical points of the area function. We can do this by taking the derivative of the area function with respect to \( l \) and setting it equal to zero. Let's calculate the derivative of the area function with respect to \( l \) and solve for the critical points. Evaluate the derivative by following steps: - step0: Evaluate the derivative: \(\frac{d}{dl}\left(l\left(1900-2l\right)\right)\) - step1: Simplify: \(\frac{d}{dl}\left(1900l-2l^{2}\right)\) - step2: Use differentiation rules: \(\frac{d}{dl}\left(1900l\right)-\frac{d}{dl}\left(2l^{2}\right)\) - step3: Calculate: \(1900-4l\) The derivative of the area function with respect to \( l \) is \( 1900 - 4l \). To find the critical points, we set the derivative equal to zero and solve for \( l \): \[ 1900 - 4l = 0 \] \[ 4l = 1900 \] \[ l = \frac{1900}{4} \] \[ l = 475 \] So, the critical point occurs at \( l = 475 \) meters. Now, we need to check if this critical point corresponds to a maximum area. We can do this by checking the second derivative of the area function or by substituting the critical point back into the area function. Let's calculate the second derivative of the area function with respect to \( l \) to confirm that the critical point corresponds to a maximum area. Evaluate the derivative by following steps: - step0: Evaluate the derivative: \(\frac{d}{dl}\left(1900-4l\right)\) - step1: Use differentiation rules: \(\frac{d}{dl}\left(1900\right)-\frac{d}{dl}\left(4l\right)\) - step2: Find the derivative: \(0-4\) - step3: Remove 0: \(-4\) The second derivative of the area function with respect to \( l \) is -4. Since the second derivative is negative, the critical point at \( l = 475 \) meters corresponds to a maximum area. Now, we can find the dimensions of the rectangular plot that encloses the largest area. The length of the side parallel to the river is 475 meters, and the width of the side perpendicular to the river is: \[ w = 1900 - 2 \times 475 = 950 \text{ meters} \] Therefore, the dimensions of the rectangular plot that encloses the largest area are 475 meters by 950 meters.