A rancher has 140 feet of fencing to enclose three sides of a rectangular garden; the fourth side is a cliff wall and doesn't need fencing. Determine the dimensions of the garden that will maximize the area. In this problem, the width is the smaller dimension, requiring two sides of fencing, while the length is the larger dimension, requiring only one side of fencing.

To maximize the area of the rectangular garden with three sides to be fenced, we need to use calculus to find the dimensions that will give the largest area.

Let's denote the width of the garden by \( w \) and the length of the garden by \( l \). Since the width requires fencing on both sides, we have \( 2w + l = 140 \) feet of fencing available. The area \( A \) of the rectangle is given by the product of the length and width: \( A = w \times l \).

From the fencing constraint \( 2w + l = 140 \), we can express \( l \) in terms of \( w \): \( l = 140 - 2w \).

Now we substitute \( l \) into the area formula: \( A(w) = w \times (140 - 2w) \) \( A(w) = 140w - 2w^2 \)

To maximize this area, we need to find the critical points by taking the derivative of \( A(w) \) with respect to \( w \) and setting it equal to zero. \( \frac{dA}{dw} = 140 - 4w \)

Setting the derivative equal to zero to find the critical points: \( 140 - 4w = 0 \) \( 4w = 140 \) \( w = \frac{140}{4} \) \( w = 35 \)

We found that the width \( w \) that maximizes the area is 35 feet. Substituting \( w = 35 \) back into the constraint equation to find \( l \): \( l = 140 - 2(35) \) \( l = 140 - 70 \) \( l = 70 \)

Therefore, the width of the garden is 35 feet, and the length is 70 feet. These dimensions will maximize the area of the garden.

To ensure this is the maximum, one would typically take the second derivative of A(w) and show that it's negative at \( w = 35 \), confirming that this point yields a maximum.

Extra: The problem involves an application of optimization in calculus. Optimization is the process of finding the best or most efficient use of resources within a given set of constraints. The task typically involves maximizing or minimizing some function. In this case, we aimed to maximize the area of a rectangle given a fixed perimeter (or a fixed length of fencing).

We assigned variables to the dimensions of the rectangle, set up an equation for the perimeter involving the dimensions of the garden, and expressed one variable in terms of the other to get an area function of one variable, \( A(w) \), that we can then optimize.

The optimization process involves finding the derivative of the area function with respect to the width, setting this derivative equal to zero to find critical points, and then determining which of these points actually give the maximum area by testing or using the second derivative. When the second derivative is negative, the function is concave down, indicating a maximum at that point.

This approach is common in many real-world problems where resources are limited, and an optimal solution is sought within those constraints. It also provides a good illustration of how calculus can be applied to practical and tangible problems.