To find the dimensions of a rectangular corral that produces the greatest enclosed area given 200 feet of fencing, we will use optimization techniques.

Problem Summary:

• Total perimeter of the rectangular corral: 200 feet.
• Let the length of the corral be $L$ and the width be $W$.
• We need to maximize the area $A$ of the rectangle.

Step 1: Write the perimeter constraint.

The perimeter $P$ of a rectangle is given by: $P = 2L + 2W$ Given that $P = 200$: $2L + 2W = 200$ Dividing through by 2: $L + W = 100$

Step 2: Write the area function.

The area $A$ of the rectangle is: $A = L \times W$

Step 3: Express $A$ in terms of one variable.

From the perimeter constraint $L + W = 100$, we can express $W$ in terms of $L$: $W = 100 - L$ Substitute this into the area function: $A = L \times (100 - L)$ $A = 100L - L^2$

Step 4: Find the critical points.

To find the value of $L$ that maximizes $A$, take the derivative of $A$ with respect to $L$ and set it equal to zero: $\frac{dA}{dL} = 100 - 2L = 0$ $2L = 100$ $L = 50$

Step 5: Determine $W$.

Using $W = 100 - L$: $W = 100 - 50 = 50$

Step 6: Verify that the solution gives a maximum.

Since the area function $A = 100L - L^2$ is a quadratic function that opens downward (because the coefficient of $L^2$ is negative), the critical point at $L = 50$ gives a maximum.

Final Answer:

The dimensions of the rectangular corral that produce the greatest enclosed area are:

• Length: 50 feet
• Width: 50 feet

The enclosed area is a square with side lengths of 50 feet.

Would you like a detailed explanation of the derivative process or more information about quadratic optimization?

Here are 5 related questions:

1. How would the solution change if the fencing was 300 feet instead of 200 feet?
2. What if the corral needs to have a length twice as long as its width?
3. How would you maximize the area of a rectangular corral with three sides (one side open)?
4. How does the quadratic nature of the area function ensure a maximum?
5. What is the difference between finding a maximum and a minimum in calculus?

Tip: For optimization problems, always start by clearly defining your constraint and objective functions.