Let’s analyze the problem step by step.

Problem Restatement:

1. A rectangular pen is built with $1200 \, \text{m}$ of fencing.
2. The pen is divided into three parts by two parallel partitions.
3. We need to:
   • (a) Find the maximum possible area of the pen.
   • (b) Determine how the maximum area changes if each side of the pen must be at least $180 \, \text{m}$ long.

Step 1: Define variables and constraints

Let:

• $L$ = Length of the pen.
• $W$ = Width of the pen.

The total fencing is used for:

• Two lengths ($2L$).
• Four widths ($4W$) since two partitions divide the pen.

Thus, the total fencing constraint is: $2L + 4W = 1200.$ Simplify: L + 2W = 600. \tag{1}

The area of the pen is: A = L \cdot W. \tag{2}

Step 2: Solve for $L$ in terms of $W$

From equation (1): L = 600 - 2W. \tag{3}

Substitute equation (3) into equation (2): $A = (600 - 2W) \cdot W.$ Simplify: A = 600W - 2W^2. \tag{4}

Step 3: Maximize the area $A$

To find the maximum area, take the derivative of $A$ with respect to $W$ and set it to zero: $\frac{dA}{dW} = 600 - 4W = 0.$ Solve for $W$: $W = 150 \, \text{m}.$

Substitute $W = 150$ into equation (3) to find $L$: $L = 600 - 2(150) = 300 \, \text{m}.$

Thus, the dimensions for maximum area are: $L = 300 \, \text{m}, \quad W = 150 \, \text{m}.$

The maximum area is: $A = L \cdot W = 300 \cdot 150 = 45,000 \, \text{m}^2.$

Step 4: Address part (b) - Constraints on dimensions

If both $L \geq 180 \, \text{m}$ and $W \geq 180 \, \text{m}$, the constraint limits the possible values of $W$. From equation (1), $L = 600 - 2W$, so: $600 - 2W \geq 180 \quad \Rightarrow \quad W \leq 210.$

Thus: $180 \leq W \leq 210.$

Recalculate the area $A$ for $W = 180$ and $W = 210$:

• If $W = 180$: $L = 600 - 2(180) = 240, \quad A = 240 \cdot 180 = 43,200 \, \text{m}^2.$
• If $W = 210$: $L = 600 - 2(210) = 180, \quad A = 180 \cdot 210 = 37,800 \, \text{m}^2.$

The maximum area within these constraints is: 43,200 \, \text{m}^2 \, \text{(at W = 180, L = 240)}.

Final Answers:

(a) The maximum possible area is: 45,000 \, \text{m}^2 \, \text{(at L = 300, W = 150)}.

(b) If $L \geq 180 \, \text{m}$ and $W \geq 180 \, \text{m}$, the maximum area decreases to: 43,200 \, \text{m}^2 \, \text{(at L = 240, W = 180)}.

Would you like further clarification or a detailed graph of the solutions? Here are 5 related questions to expand on this:

1. What is the effect of increasing the total fencing on the maximum area?
2. How do we confirm that the solution represents a maximum and not a minimum?
3. Can we optimize the pen area if partitions are not required?
4. What happens if we change the number of partitions (e.g., 1 or 3 partitions)?
5. Can we solve this using numerical methods or graphing software?

Tip: When maximizing or minimizing areas with constraints, always express one variable in terms of the other using the given constraints.