Farmer Brown has 610 yards of fencing with which to build a rectangular corral. He builds a new pen that uses the river as one side, so he only has to fence the other 3 sides (see figure below).

Answer the following:

a) Write the area of the corral as a function of x.

A(x) =

b) Determine the maximum area enclosed by the corral.Maximum area =  square yards.

We are tasked with determining the maximum area for a rectangular corral built by Farmer Brown, with one side along a river (so only 3 sides need fencing) and 610 yards of fencing available.

### Step 1: Define the Variables
Let:
- \( x \) be the length of the corral parallel to the river (so this side doesn't need fencing).
- \( y \) be the width of the corral perpendicular to the river (these two sides need fencing).

The total amount of fencing available is 610 yards. Since only 3 sides need to be fenced (one length \( x \) and two widths \( y \)), the perimeter equation is:

\[
x + 2y = 610
\]

### Step 2: Express \( y \) in terms of \( x \)
Solving the perimeter equation for \( y \):

\[
2y = 610 - x
\]

\[
y = \frac{610 - x}{2}
\]

### Step 3: Write the Area Function in Terms of \( x \)
The area \( A \) of the corral is the length times the width:

\[
A(x) = x \cdot y
\]

Substitute the expression for \( y \) from Step 2:

\[
A(x) = x \cdot \frac{610 - x}{2}
\]

Simplify the expression:

\[
A(x) = \frac{x(610 - x)}{2}
\]

\[
A(x) = \frac{610x - x^2}{2}
\]

Thus, the area of the corral as a function of \( x \) is:

\[
A(x) = \frac{610x - x^2}{2}
\]

### Step 4: Maximize the Area
To find the maximum area, we need to find the critical points by taking the derivative of \( A(x) \) with respect to \( x \) and setting it equal to zero.

The area function is:

\[
A(x) = \frac{610x - x^2}{2}
\]

The derivative of \( A(x) \) is:

\[
A'(x) = \frac{1}{2} \cdot (610 - 2x)
\]

Set \( A'(x) = 0 \) to find the critical points:

\[
\frac{1}{2}(610 - 2x) = 0
\]

\[
610 - 2x = 0
\]

\[
2x = 610
\]

\[
x = 305
\]

### Step 5: Determine the Maximum Area
Now, substitute \( x = 305 \) back into the expression for \( y \):

\[
y = \frac{610 - 305}{2} = \frac{305}{2} = 152.5
\]

Thus, the dimensions of the corral that maximize the area are:
- \( x = 305 \) yards (length along the river),
- \( y = 152.5 \) yards (width perpendicular to the river).

The maximum area is:

\[
A(305) = \frac{610 \times 305 - 305^2}{2}
\]

\[
A(305) = \frac{186050 - 93025}{2}
\]

\[
A(305) = \frac{93025}{2} = 46512.5 \text{ square yards}
\]

### Final Answers:
a) The area of the corral as a function of \( x \) is:

\[
A(x) = \frac{610x - x^2}{2}
\]

b) The maximum area enclosed by the corral is:

\[
46512.5 \text{ square yards}
\]

Would you like further details or clarifications?

Here are some related questions for practice:

1. How would the maximum area change if 800 yards of fencing were available?
2. What would happen to the dimensions of the corral if the fencing were halved?
3. If the corral had to be square-shaped, what would its maximum area be with 610 yards of fencing?
4. What is the second derivative of the area function, and how does it confirm the maximum?
5. How would the problem change if the river required an additional buffer of 10 yards along the corral?

**Tip:** When optimizing functions, the second derivative test can confirm whether a critical point is a maximum or minimum.

Farmer Brown has 610 yards of fencing with which to build a rectangular corral. He builds a new pen that uses the river as one side, so he only has to fence the other 3 sides. Write the area of the corral as a function of x. Determine the maximum area enclosed by the corral.

Solve Farmer Brown's fencing problem using algebraic optimization. Learn how to write the area as a function of x, determine critical points, and maximize the enclosed area. Suitable for high school students learning quadratic equations and optimization techniques.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation