A farmer wants to build a rectangular pen along a straight river. She wants to divide the pen
into 5 equal rectangular pieces as shown in the picture. She will not fence one of the long sides along the river.
(a) What is the largest area she can enclose with 3,000 feet of fencing?
(b) What are the length and width of the individual pens that will give the largest area?



This problem involves maximizing the area of a rectangular pen that is divided into five equal pieces, without fencing along one side of the rectangle (since the river acts as a natural boundary).

Let’s break this down and solve it step by step.

### Step 1: Define the variables
- Let \( L \) be the length of the pen (the side parallel to the river).
- Let \( W \) be the width of the pen (the side perpendicular to the river).
  
Since the pen is divided into 5 smaller rectangular sections, there are 4 vertical divisions required along the width.

### Step 2: Set up the equation for the fencing
The total length of fencing available is 3,000 feet. The fencing is used for:
- One side of length \( L \) (opposite to the river),
- Two sides of width \( W \) (on the left and right),
- Four internal vertical dividers along the width.

The total length of the fencing can therefore be expressed as:
\[
\text{Total Fencing} = L + 6W = 3000 \, \text{feet}
\]

This gives us the constraint:
\[
L + 6W = 3000
\]

### Step 3: Express \( L \) in terms of \( W \)
We solve for \( L \) in terms of \( W \):
\[
L = 3000 - 6W
\]

### Step 4: Set up the area function
The area of the entire pen is given by:
\[
A = L \times W
\]
Substituting \( L = 3000 - 6W \) into the area formula:
\[
A(W) = (3000 - 6W) \times W = 3000W - 6W^2
\]

### Step 5: Maximize the area function
To find the value of \( W \) that maximizes the area, we take the derivative of \( A(W) \) with respect to \( W \) and set it equal to zero.

The derivative of \( A(W) \) is:
\[
\frac{dA}{dW} = 3000 - 12W
\]
Setting this equal to zero:
\[
3000 - 12W = 0
\]
\[
W = 250 \, \text{feet}
\]

### Step 6: Find \( L \)
Now, substitute \( W = 250 \) feet into the equation \( L = 3000 - 6W \):
\[
L = 3000 - 6(250) = 3000 - 1500 = 1500 \, \text{feet}
\]

### Step 7: Calculate the maximum area
The maximum area is:
\[
A = L \times W = 1500 \times 250 = 375,000 \, \text{square feet}
\]

### Final Answer:
(a) The largest area the farmer can enclose is **375,000 square feet**.

(b) The dimensions of the individual pens that will give the largest area are:
- **Length** of each pen: \( L = 1500 \, \text{feet} \)
- **Width** of each pen: \( W = 250 \, \text{feet} \)

### Questions:
1. Why does maximizing the width result in maximizing the overall area?
2. How would the solution change if more than 3,000 feet of fencing were available?
3. Could the farmer enclose a larger area by changing the number of divisions (e.g., 4 pens instead of 5)?
4. What happens to the area if the fencing along the river was required?
5. How would the solution change if the river wasn't straight?

### Tip:
Always express one variable in terms of the other when dealing with constrained optimization problems to reduce the number of variables in the equation.

A farmer wants to build a rectangular pen along a straight river. She wants to divide the pen into 5 equal rectangular pieces as shown in the picture. She will not fence one of the long sides along the river. (a) What is the largest area she can enclose with 3,000 feet of fencing? (b) What are the length and width of the individual pens that will give the largest area?

Discover how to maximize the area of a rectangular pen divided into 5 sections with 3,000 feet of fencing along a river. Learn key optimization concepts, algebraic techniques, and differentiation to find the largest possible area. This problem is perfect for students in Grades 11-12 studying quadratic functions and optimization.

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