A rancher is planning to build a rectangular fence around his property, with one side of the property being next to a cliff. The property is 6150 square yards, and only the three sides not adjacent to the cliff will be fenced off.

What is the least amount of fencing required for this job?

Let \( x \) represent the width of the rectangular fenced area (the side perpendicular to the cliff), and let \( y \) represent the length (parallel to the cliff). Since the area of the fenced region is given as 6150 square yards, we have:

\[
x \cdot y = 6150
\]

The fencing will cover the two widths and one length (only three sides), so the total length of fencing needed, \( F \), is:

\[
F = 2x + y
\]

Our goal is to minimize \( F \), given the constraint \( x \cdot y = 6150 \). To solve this, we can express \( y \) in terms of \( x \) using the area equation:

\[
y = \frac{6150}{x}
\]

Substitute this expression for \( y \) in the fencing equation:

\[
F = 2x + \frac{6150}{x}
\]

Now, we have a function of \( x \) alone:

\[
F(x) = 2x + \frac{6150}{x}
\]

To find the minimum fencing required, we’ll take the derivative of \( F(x) \) with respect to \( x \), set it equal to zero, and solve for \( x \).

1. **Differentiate \( F(x) \) with respect to \( x \):**

   \[
   F'(x) = 2 - \frac{6150}{x^2}
   \]

2. **Set the derivative equal to zero to find critical points:**

   \[
   2 - \frac{6150}{x^2} = 0
   \]

   Rearrange to solve for \( x \):

   \[
   \frac{6150}{x^2} = 2
   \]

   \[
   x^2 = \frac{6150}{2} = 3075
   \]

   \[
   x = \sqrt{3075} \approx 55.46
   \]

3. **Find \( y \) using \( y = \frac{6150}{x} \):**

   \[
   y = \frac{6150}{55.46} \approx 110.92
   \]

4. **Calculate the minimum fencing \( F \):**

   \[
   F = 2x + y = 2(55.46) + 110.92 = 110.92 + 110.92 = 221.84
   \]

Thus, the least amount of fencing required is approximately **221.84 yards**.

Would you like further details or have any questions?

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**Follow-up Questions:**

1. How would the solution change if all four sides were fenced?
2. What if the area was increased to 10,000 square yards?
3. How does this optimization apply in a real-world ranch setting?
4. Could we solve this problem using different variables for width and length?
5. How does calculus help in optimizing resource use in general?

**Tip:** Optimization problems often benefit from identifying constraints first and expressing all variables in terms of one another to simplify.

A rancher is planning to build a rectangular fence around his property, with one side of the property being next to a cliff. The property is 6150 square yards, and only the three sides not adjacent to the cliff will be fenced off. What is the least amount of fencing required for this job?

Discover how to minimize the fencing required for a ranch property adjacent to a cliff with an area of 6150 square yards. This problem involves key concepts in calculus and optimization, leading to a minimum fencing requirement of approximately 221.84 yards.

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