You have 550 feet of fencing to enclose a rectangular plot that borders on a river. If you do not fence the side along the river, find the length and width of the plot that will maximize the area. What is the largest area that can be enclosed?

To maximize the area of a rectangular plot that borders a river, we can set up the problem as follows: Let: - \( L \) be the length of the side parallel to the river. - \( W \) be the width of the side perpendicular to the river. Since the side along the river does not require fencing, the total amount of fencing used for the other three sides is given by: \[ L + 2W = 550 \] From this equation, we can express \( L \) in terms of \( W \): \[ L = 550 - 2W \] The area \( A \) of the rectangle can be expressed as: \[ A = L \times W = (550 - 2W) \times W \] Expanding this, we get: \[ A = 550W - 2W^2 \] This is a quadratic equation in the standard form \( A = -2W^2 + 550W \). To find the maximum area, we can use the vertex formula for a parabola, which is given by \( W = -\frac{b}{2a} \), where \( a \) and \( b \) are the coefficients from the quadratic equation \( A = aW^2 + bW + c \). Here, \( a = -2 \) and \( b = 550 \): \[ W = -\frac{550}{2 \times -2} = \frac{550}{4} = 137.5 \] Now, substituting \( W = 137.5 \) back into the equation for \( L \): \[ L = 550 - 2(137.5) = 550 - 275 = 275 \] Thus, the dimensions that maximize the area are: - Length \( L = 275 \) feet - Width \( W = 137.5 \) feet Now, we can calculate the maximum area: \[ A = L \times W = 275 \times 137.5 \] Calculating this gives: \[ A = 275 \times 137.5 = 37812.5 \text{ square feet} \] Therefore, the dimensions that maximize the area are: - Length: 275 feet - Width: 137.5 feet And the largest area that can be enclosed is: \[ \boxed{37812.5} \text{ square feet} \]