The back of Dante's property is a creek. Dante would like to enclose a rectangular area, using the creek as one side and fencing for the other three sides, to create a corral. If there is 200
feet of fencing available, what is the maximum possible area of the corral?

To solve this problem, we need to maximize the area of a rectangular corral with three sides fenced, using the creek as the fourth side. Let's define the variables and set up the equations:

1. **Variables**:
   - \( x \): the length of the side perpendicular to the creek.
   - \( y \): the length of the side parallel to the creek.

2. **Fencing constraint**:
   Since the creek acts as one side, the fencing is only needed for the three other sides (two sides of length \( x \) and one side of length \( y \)). The total amount of fencing available is 200 feet, so we have:
   \[
   2x + y = 200
   \]

3. **Area of the corral**:
   The area \( A \) of the corral is given by:
   \[
   A = x \times y
   \]

To find the maximum possible area, we need to express \( A \) in terms of a single variable. Let's solve for \( y \) from the fencing constraint:

\[
y = 200 - 2x
\]

Substitute \( y \) into the area formula:

\[
A(x) = x(200 - 2x)
\]

Simplify this equation:

\[
A(x) = 200x - 2x^2
\]

Now, we have a quadratic function \( A(x) = -2x^2 + 200x \). This quadratic function opens downwards (since the coefficient of \( x^2 \) is negative), so its maximum value occurs at its vertex.

The x-coordinate of the vertex of a quadratic function \( ax^2 + bx + c \) is given by:

\[
x = -\frac{b}{2a}
\]

For our function, \( a = -2 \) and \( b = 200 \):

\[
x = -\frac{200}{2(-2)} = \frac{200}{4} = 50
\]

Now, substitute \( x = 50 \) back into the equation for \( y \):

\[
y = 200 - 2(50) = 200 - 100 = 100
\]

Thus, the dimensions that maximize the area are \( x = 50 \) feet and \( y = 100 \) feet.

Finally, the maximum possible area \( A \) is:

\[
A = x \times y = 50 \times 100 = 5000 \text{ square feet}
\]

### Conclusion

The maximum possible area of the corral that Dante can enclose with 200 feet of fencing, using the creek as one side, is **5000 square feet**.

Would you like more details on any part of the solution, or do you have any other questions?

### Related Questions:

1. How would the area change if the amount of fencing available were 250 feet?
2. What if the creek is not straight but forms a semicircle around one side of the corral? How would this change the problem?
3. How can you determine if a quadratic function opens upwards or downwards?
4. If the fencing is unevenly distributed (e.g., different lengths for the sides perpendicular to the creek), how would you maximize the area?
5. What is the significance of the vertex in the context of this problem?

### Tip:

When dealing with optimization problems, always identify the constraints and express the quantity to be optimized in terms of a single variable to simplify the problem.

Learn how to maximize the area of a rectangular corral using a creek as one side and 200 feet of fencing for the other sides. Step-by-step solution includes setting up equations, solving quadratic functions, and optimizing area. Suitable for Grades 9-12 students.

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