Part 1a: Maximizing the Area of the Pen

Problem Setup:

• One side of the pen is along the barn.
• The total fencing used for the other three sides is $500 \, \text{m}$.
• Let:
  • $x$ = length of each side perpendicular to the barn (2 such sides).
  • $y$ = length of the side parallel to the barn.

The total fencing constraint is: $2x + y = 500.$

Objective:

Maximize the area of the pen: $A = x \cdot y.$

Step 1: Express $y$ in terms of $x$:

From the fencing constraint: $y = 500 - 2x.$

Step 2: Substitute $y$ into the area formula:

$A = x \cdot (500 - 2x).$ Simplify: $A = 500x - 2x^2.$

Step 3: Maximize $A$ using calculus:

The derivative of $A$ is: $\frac{dA}{dx} = 500 - 4x.$ Set $\frac{dA}{dx} = 0$ to find critical points: $500 - 4x = 0 \implies x = 125.$

Step 4: Verify maximum using the second derivative:

The second derivative is: $\frac{d^2A}{dx^2} = -4,$ which is negative, confirming a maximum.

Step 5: Calculate $y$:

$y = 500 - 2(125) = 250.$

Final Dimensions:

The dimensions that maximize the area are: $x = 125 \, \text{m} \quad (\text{perpendicular to the barn}),$ $y = 250 \, \text{m} \quad (\text{parallel to the barn}).$

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Part 1b: Minimizing Fencing for Four Pens

Problem Setup:

• Four identical adjacent pens are constructed along a barn.
• Total area of all pens = $4 \cdot 225 = 900 \, \text{m}^2$.
• Let:
  • $x$ = width of each pen (perpendicular to the barn).
  • $y$ = length of each pen (parallel to the barn).

Objective:

Minimize the fencing:

• The total fencing includes:
  • 5 vertical segments ($5x$): 4 internal and 1 external width.
  • 1 horizontal segment ($y$) for the barn-side.
  • 4 horizontal segments ($4y$) opposite the barn.

Thus, the fencing required is: $F = 5x + 5y.$

Step 1: Relate $x$ and $y$ using area:

Each pen's area is $xy = 225$: $y = \frac{225}{x}.$

Step 2: Substitute $y$ into $F$:

$F = 5x + 5\left(\frac{225}{x}\right).$ Simplify: $F = 5x + \frac{1125}{x}.$

Step 3: Minimize $F$ using calculus:

The derivative of $F$ is: $\frac{dF}{dx} = 5 - \frac{1125}{x^2}.$ Set $\frac{dF}{dx} = 0$ to find critical points: $5 = \frac{1125}{x^2} \implies x^2 = 225 \implies x = 15.$

Step 4: Calculate $y$:

$y = \frac{225}{x} = \frac{225}{15} = 15.$

Final Dimensions:

Each pen has: $x = 15 \, \text{m} \quad (\text{width}),$ $y = 15 \, \text{m} \quad (\text{length}).$

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Summary:

1. Part 1a: Dimensions to maximize the area:
   • $x = 125 \, \text{m}$,
   • $y = 250 \, \text{m}$.
2. Part 1b: Dimensions to minimize fencing:
   • $x = 15 \, \text{m}$,
   • $y = 15 \, \text{m}$.

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Would you like detailed explanations for any specific steps?

Related Questions:

1. Why does calculus ensure the optimal solution in these problems?
2. How would the solution change if more pens were added in Part 1b?
3. What if the barn were not present in Part 1a—how would the dimensions change?
4. Can you show a graph of the area and fencing functions to visualize the optimization?
5. What is the role of symmetry in minimizing the fencing in Part 1b?

Tip:

Always verify optimization solutions by checking both the first and second derivatives to confirm the nature of the critical points.