In this problem, we need to maximize the area of a rectangular swimming area with a rope and floats, with one side of the rectangle being the shoreline. We are given that the total amount of rope available for the other three sides of the rectangle is 2300 yards.

Step 1: Define Variables

Let:

• $L$ be the length of the rectangle parallel to the shoreline.
• $W$ be the width of the rectangle (the other sides perpendicular to the shoreline).

Step 2: Set Up the Constraint

Since the shoreline forms one side of the rectangle, only three sides need to be roped off: two widths and one length. The total length of the rope is used for these three sides, so we can write the constraint as:

$2W + L = 2300$

Step 3: Express the Area

The area $A$ of the rectangle is given by:

$A = L \times W$

Step 4: Solve for One Variable

From the constraint $2W + L = 2300$, solve for $L$:

$L = 2300 - 2W$

Step 5: Substitute into the Area Formula

Substitute $L = 2300 - 2W$ into the area formula $A = L \times W$:

$A(W) = (2300 - 2W) \times W$ $A(W) = 2300W - 2W^2$

Step 6: Maximize the Area

To maximize the area, we take the derivative of $A(W)$ with respect to $W$, and set it equal to zero.

$\frac{dA}{dW} = 2300 - 4W$

Set the derivative equal to zero to find the critical point:

$2300 - 4W = 0$ $4W = 2300$ $W = 575$

Step 7: Verify Maximum

To confirm that this value of $W$ gives a maximum, check the second derivative of $A(W)$:

$\frac{d^2A}{dW^2} = -4$

Since the second derivative is negative, the function is concave down, meaning $W = 575$ is a maximum.

Step 8: Find the Corresponding Length $L$

Now that we know $W = 575$, substitute this value back into the constraint to find $L$:

$L = 2300 - 2(575) = 2300 - 1150 = 1150$

Step 9: Calculate the Maximum Area

The maximum area is:

$A = L \times W = 1150 \times 575 = 661250 \text{ square yards}$

Final Answer:

• The dimensions that maximize the area are:
  • Length $L = 1150$ yards
  • Width $W = 575$ yards
• The maximum area is $661,250$ square yards.

Would you like a more detailed explanation of any step, or have additional questions?

Related Questions:

1. How would the maximum area change if the total length of the rope were increased to 3000 yards?
2. What is the formula for the area of a rectangle, and how is it used to solve optimization problems like this one?
3. How can calculus be used to optimize real-world problems like maximizing area with constraints?
4. What are some practical considerations when using a fixed-length rope to enclose a rectangular area?
5. What would happen if we used a circular enclosure instead of a rectangle, and how could we optimize that?

Tip: When maximizing or minimizing a function, always check the second derivative to ensure the solution