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.