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To solve this problem, let’s go through it step-by-step.

### Given:
- A rectangular field with **three walls of fencing**, with the fourth side against the gym wall.
- The **area** of the field is $ 360,000 \, \text{ft}^2 $.
- The **cost of fencing is $89 per foot**.

### Goal:
Minimize the fencing cost while maintaining the area at 360,000 ft².

### Step 1: Define Variables
Let:
- $ x $ = length of the side parallel to the gym wall (no fencing needed here).
- $ y $ = width of the field, which will have fencing.

### Step 2: Set Up the Area Equation
The area of the rectangle is given by:
\[
x \cdot y = 360,000
\]
Thus, we can express $ x $ in terms of $ y $:
\[
x = \frac{360,000}{y}
\]

### Step 3: Set Up the Fencing Cost Function
Since only three sides need fencing (two widths $ y $ and one length $ x $), the total length of fencing required is:
\[
\text{Total Fencing} = x + 2y
\]
Substitute $ x $ from the area equation:
\[
\text{Total Fencing} = \frac{360,000}{y} + 2y
\]

### Step 4: Formulate the Cost Function
The cost $ C $ in dollars is:
\[
C = 89 \cdot \left( \frac{360,000}{y} + 2y \right)
\]
Expanding this:
\[
C = 89 \cdot \frac{360,000}{y} + 178y
\]

### Step 5: Differentiate the Cost Function with Respect to $ y $ and Find Critical Points
To minimize $ C $, we take the derivative of $ C $ with respect to $ y $ and set it equal to zero.

1. Differentiate $ C $ with respect to $ y $:
   \[
   \frac{dC}{dy} = -\frac{89 \cdot 360,000}{y^2} + 178
   \]
2. Set $ \frac{dC}{dy} = 0 $ to find the critical point:
   \[
   -\frac{89 \cdot 360,000}{y^2} + 178 = 0
   \]
   Solving for $ y $:
   \[
   \frac{89 \cdot 360,000}{y^2} = 178
   \]
   \[
   y^2 = \frac{89 \cdot 360,000}{178}
   \]
   \[
   y^2 = 180,000
   \]
   \[
   y = \sqrt{180,000} = 424.26 \, \text{ft}
   \]

3. Substitute $ y = 424.26 $ back into the area equation to find $ x $:
   \[
   x = \frac{360,000}{424.26} \approx 848.53 \, \text{ft}
   \]

### Step 6: Calculate the Total Cost
Now that we have the dimensions:
- $ x = 848.53 $ ft
- $ y = 424.26 $ ft

The total fencing length needed is:
\[
\text{Total Fencing} = x + 2y = 848.53 + 2 \cdot 424.26 = 1697.05 \, \text{ft}
\]

The cost is:
\[
C = 89 \cdot 1697.05 \approx 151,438.45
\]

### Final Answer:
The least amount of money needed is **$151,438.45**.

Would you like further details on any of these steps, or have any questions?

---

#### Related Questions:
1. What is the general method for minimizing costs in fencing problems?
2. How would the cost change if the price per foot of fencing increased?
3. What if the field area were different? How would that affect the dimensions?
4. Why do we only differentiate with respect to $ y $ in this problem?
5. How can we verify that this solution gives the minimum cost?

#### Tip:
In optimization problems, always check if critical points provide a minimum or maximum by evaluating endpoints or using the second derivative test.

A rectangular recreational field needs to be built outside of a gymnasium. Three walls of fencing are needed, and the fourth wall is to be a wall of the gymnasium itself. The ideal area for such a field is exactly 360,000 ft². To minimize costs, it is necessary to construct the fencing using the least amount of material possible. Assuming that the material used in the fencing costs $89/ft, what is the least amount of money needed to build this fence of ideal area? Round your answer to the nearest two decimal places.

Learn how to calculate the minimum cost to build a fence around a rectangular field with an area of 360,000 ft². This problem covers optimization and cost minimization techniques using calculus, suitable for students in Grades 10-12.

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