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The problem you're solving involves finding the dimensions of five identical pens arranged side by side, where the total area is \( 1000 \, m^2 \), and we need to minimize the amount of fencing used. Here's how we approach it:

### 1. Area Equation:
The total area of the five pens is given by:
\[
1000 = 5xy
\]
where \(x\) is the length of the pens and \(y\) is the width of each pen. This equation can be rewritten to solve for \(x\):
\[
x = \frac{1000}{5y} = \frac{200}{y}
\]

### 2. Fence Equation:
The total length of fencing required (perimeter) is given by:
\[
F = 6x + 10y
\]
This includes the lengths of the horizontal sections (6 \(x\), as there are six vertical boundaries) and the vertical sections (10 \(y\), as there are five pens requiring five boundaries and two external ones).

### 3. Substituting \(x\) into the Fence Equation:
Now substitute \(x = \frac{200}{y}\) into the equation for the fence:
\[
F = 6 \left( \frac{200}{y} \right) + 10y = \frac{1200}{y} + 10y
\]

### 4. Minimizing the Fence Equation:
To minimize the amount of fence used, take the derivative of \(F\) with respect to \(y\) and set it equal to zero:
\[
\frac{dF}{dy} = -\frac{1200}{y^2} + 10
\]
Setting this equal to zero:
\[
-\frac{1200}{y^2} + 10 = 0
\]
\[
\frac{1200}{y^2} = 10
\]
\[
y^2 = \frac{1200}{10} = 120
\]
\[
y = \sqrt{120} \approx 10.95 \, m
\]

### 5. Finding \(x\):
Now, substitute \(y \approx 10.95\) back into the equation for \(x\):
\[
x = \frac{200}{y} = \frac{200}{10.95} \approx 18.27 \, m
\]

### Final Dimensions:
- Width \(y \approx 10.95 \, m\)
- Length \(x \approx 18.27 \, m\)

These dimensions will minimize the amount of fencing used while keeping the total area of the pens at \(1000 \, m^2\).

Would you like more details or have any questions?

### Related Questions:
1. How do we confirm that the calculated dimensions give the correct total area of \(1000 \, m^2\)?
2. Can we use Lagrange multipliers to solve this optimization problem?
3. What happens to the fence equation if the number of pens changes from 5 to another value?
4. What if the area constraint changes to something other than \(1000 \, m^2\)?
5. How does the shape of the pens (square vs rectangular) affect the fencing requirements?

### Tip:
When optimizing a quantity, always check if the second derivative confirms a minimum or maximum!

You’re building five identical pens adjacent to each other with a total area of 1000 m². What dimensions should you use to minimize the amount of fence?

Discover how to minimize the amount of fencing needed for five identical pens with a total area of 1000 m². Learn about the application of optimization and calculus in solving real-world geometry problems. Suitable for advanced high school and college-level students.

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