To determine the dimensions of the rectangular grazing area that require the least amount of fencing, given the conditions in the problem, let’s solve it step by step.

1. Define Variables:

• $x$: the length of the grazing area perpendicular to the existing fence.
• $y$: the width of the grazing area parallel to the existing fence.
• The area of the grazing field is 180,000 square yards, so: $x \cdot y = 180,000.$
• The farmer needs to fence three sides of the grazing area (2 widths and 1 length).

2. Express the Total Fencing Length:

The total fencing $P$ (perimeter for 3 sides) is given by: $P = x + 2y.$ Our goal is to minimize $P$ while satisfying the area constraint $x \cdot y = 180,000$.

3. Substitute $x$ in Terms of $y$:

From the area constraint $x \cdot y = 180,000$, solve for $x$: $x = \frac{180,000}{y}.$ Substitute this into the equation for $P$: $P = \frac{180,000}{y} + 2y.$

4. Minimize $P$ Using Calculus:

To minimize $P$, take the derivative of $P$ with respect to $y$ and set it equal to zero: $P'(y) = \frac{d}{dy} \left( \frac{180,000}{y} + 2y \right).$

1. Differentiate each term:
   • The derivative of $\frac{180,000}{y}$ is $-\frac{180,000}{y^2}$.
   • The derivative of $2y$ is $2$.

Thus: $P'(y) = -\frac{180,000}{y^2} + 2.$

2. Set $P'(y) = 0$ to find the critical points: $-\frac{180,000}{y^2} + 2 = 0.$ Rearrange: $\frac{180,000}{y^2} = 2.$ Solve for $y^2$: $y^2 = \frac{180,000}{2} = 90,000.$ Take the square root: $y = \sqrt{90,000} = 300.$

5. Solve for $x$:

Using the area constraint $x \cdot y = 180,000$ with $y = 300$: $x = \frac{180,000}{y} = \frac{180,000}{300} = 600.$

6. Verify the Solution:

• The dimensions are:
  • $x = 600$ yards (length),
  • $y = 300$ yards (width).
• Check the area: $x \cdot y = 600 \cdot 300 = 180,000 \, \text{square yards}.$
• Total fencing: $P = x + 2y = 600 + 2(300) = 600 + 600 = 1,200 \, \text{yards}.$

Final Answer:

The dimensions that require the least amount of fencing are:

• Length ($x$) = 600 yards
• Width ($y$) = 300 yards

The total fencing needed is 1,200 yards.

Follow-up Questions:

1. How does the area constraint affect the optimization process?
2. Why is calculus useful in solving optimization problems like this one?
3. How would the solution change if fencing was required on all four sides?
4. Can you generalize this solution for any required grazing area $A$?
5. What happens if the existing fence is not straight but curved?

Tip: When solving optimization problems, always start by expressing all variables in terms of one variable using the constraints provided. This simplifies the process!