Suppose that you have 840 feet of rope and want to use it to make a rectangle. What dimensions should you make your rectangle if you want to enclose the maximum possible area?

The length should be feet

The width should be feet

The total area enclosed is square feet.

Step 1 ：The problem is asking for the dimensions of a rectangle that will enclose the maximum possible area given a fixed perimeter of 840 feet. This is a problem of optimization.

Step 2 ：The perimeter of a rectangle is given by the formula \(P = 2L + 2W\), where \(L\) is the length and \(W\) is the width. In this case, we know that the perimeter is 840 feet, so we can write the equation as \(840 = 2L + 2W\).

Step 3 ：The area of a rectangle is given by the formula \(A = LW\). We want to maximize this area.

Step 4 ：To do this, we can express one of the dimensions (say, \(W\)) in terms of the other dimension and the perimeter, and then substitute this into the area formula. This will give us an equation for the area in terms of a single variable, which we can then maximize.

Step 5 ：From the perimeter equation, we can express \(W\) as \(W = (840 - 2L) / 2 = 420 - L\). Substituting this into the area formula gives \(A = L(420 - L)\).

Step 6 ：This is a quadratic equation, and its maximum value occurs at the vertex. The x-coordinate of the vertex of a parabola given by the equation \(y = ax^2 + bx + c\) is \(-b / (2a)\). In this case, \(a = -1\) and \(b = 420\), so the x-coordinate of the vertex is \(-420 / (2*-1) = 210\). This is the value of \(L\) that maximizes the area.

Step 7 ：Substituting \(L = 210\) into the equation for \(W\) gives \(W = 420 - 210 = 210\). So the rectangle should be a square with side length 210 feet.

Step 8 ：The maximum area is then \(A = 210 * 210 = 44100\) square feet.

Step 9 ：Final Answer: The length should be \(\boxed{210}\) feet, the width should be \(\boxed{210}\) feet, and the total area enclosed is \(\boxed{44100}\) square feet.