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 $240 \, \text{m}$.

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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.