An ecologist is conducting a research project on breeding pheasants in captivity. She first must construct suitable pens. She wants a rectangular area with two additional fences across its​ width, as shown in the sketch. Find the maximum area she can enclose with 960
 
m of fencing.
	
Question content area bottom
Part 1
Select the correct choice below and fill in the answer​ box(es) to complete your choice.
A.
The length of each side of the larger rectangular area is the same and measures enter your response here
 
m.
B.
The length of the shorter side of the larger rectangular area is 160
 
m and the length of the longer side of the larger rectangular area is 240
 
m.

To solve the problem, we need to maximize the area of a rectangular pen with two additional fences across its width, given that the total length of the fencing is 960 m.

### Step 1: Define variables
Let:
- \( L \) be the length of the rectangle.
- \( W \) be the width of the rectangle.

The total fencing used is the perimeter of the rectangle plus the two additional fences across the width:
\[
2L + 3W = 960
\]

### Step 2: Express \( L \) in terms of \( W \)
Rearranging the equation for \( L \):
\[
L = \frac{960 - 3W}{2}
\]

### Step 3: Express the area
The area \( A \) of the rectangle is given by:
\[
A = L \times W
\]
Substituting \( L = \frac{960 - 3W}{2} \):
\[
A = \frac{960 - 3W}{2} \cdot W
\]
Simplify:
\[
A = \frac{960W - 3W^2}{2}
\]
\[
A = 480W - \frac{3W^2}{2}
\]

### Step 4: Maximize the area
The area \( A \) is a quadratic function in terms of \( W \):
\[
A = -\frac{3}{2}W^2 + 480W
\]
This is a downward-facing parabola, so the maximum area occurs at the vertex. The width \( W \) at the vertex is given by:
\[
W = -\frac{b}{2a}
\]
Here:
- \( a = -\frac{3}{2} \)
- \( b = 480 \)

Substitute:
\[
W = -\frac{480}{2(-\frac{3}{2})} = \frac{480}{3} = 160
\]

### Step 5: Find \( L \)
Substitute \( W = 160 \) into \( L = \frac{960 - 3W}{2} \):
\[
L = \frac{960 - 3(160)}{2} = \frac{960 - 480}{2} = \frac{480}{2} = 240
\]

### Final Answer:
The length of the shorter side is \( 160 \, \text{m} \), and the length of the longer side is \( 240 \, \text{m} \).

Would you like further explanation or visual representation of this solution? 

### Related Questions:
1. What happens if the total fencing is increased or decreased?
2. How do we determine the maximum area for a pen with more or fewer internal divisions?
3. Can you verify the solution using calculus?
4. What if the area were constrained instead of the fencing length?
5. How does adding more divisions across the width affect the area?

### Tip:
Always double-check your optimization problem by substituting the calculated dimensions back into the constraints.

An ecologist is conducting a research project on breeding pheasants in captivity. She first must construct suitable pens. She wants a rectangular area with two additional fences across its width. Find the maximum area she can enclose with 960 m of fencing.

Learn how to determine the maximum area for a rectangular pen with two additional fences across its width using 960 meters of fencing. This problem involves quadratic optimization and algebraic techniques, suitable for students in Grades 9-12.

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