A farmer is building a fence to enclose a rectangular area against an existing wall, shown in the figure below. Three of the sides will require fencing and the fourth wall already exists. If the farmer has 164 feet of fencing, what is the largest area the farmer can enclose?

To maximize the area using the existing wall as one side, we can set the length parallel to the wall as \( x \) and the width as \( y \). The fencing requirement gives us the equation \( x + 2y = 164 \). Rearranging this, we have \( x = 164 - 2y \). The area \( A \) is given by \( A = x \cdot y = (164 - 2y) \cdot y = 164y - 2y^2 \). To find the maximum area, we can derive the area function, set it to zero, and solve for \( y \). The area function is a downward-opening parabola, so its vertex will provide the maximum area. Solving \( \frac{dA}{dy} = 164 - 4y = 0 \) yields \( y = 41 \). Plugging this back gives \( x = 164 - 2(41) = 82 \). Thus, the largest area is \( A = 82 \cdot 41 = 3362 \) square feet. For the final touch, the dimensions should be \( 82\) feet for the length parallel to the wall and \( 41\) feet for the width extending outwards. Now you have an area fit for a king… or, in this case, a very happy farmer!