Math Problem Statement

HELP ME SOLVE

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.

Solution

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 360,000ft2360,000 \, \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:

  • xx = length of the side parallel to the gym wall (no fencing needed here).
  • yy = width of the field, which will have fencing.

Step 2: Set Up the Area Equation

The area of the rectangle is given by: xy=360,000x \cdot y = 360,000 Thus, we can express xx in terms of yy: x=360,000yx = \frac{360,000}{y}

Step 3: Set Up the Fencing Cost Function

Since only three sides need fencing (two widths yy and one length xx), the total length of fencing required is: Total Fencing=x+2y\text{Total Fencing} = x + 2y Substitute xx from the area equation: Total Fencing=360,000y+2y\text{Total Fencing} = \frac{360,000}{y} + 2y

Step 4: Formulate the Cost Function

The cost CC in dollars is: C=89(360,000y+2y)C = 89 \cdot \left( \frac{360,000}{y} + 2y \right) Expanding this: C=89360,000y+178yC = 89 \cdot \frac{360,000}{y} + 178y

Step 5: Differentiate the Cost Function with Respect to yy and Find Critical Points

To minimize CC, we take the derivative of CC with respect to yy and set it equal to zero.

  1. Differentiate CC with respect to yy: dCdy=89360,000y2+178\frac{dC}{dy} = -\frac{89 \cdot 360,000}{y^2} + 178

  2. Set dCdy=0\frac{dC}{dy} = 0 to find the critical point: 89360,000y2+178=0-\frac{89 \cdot 360,000}{y^2} + 178 = 0 Solving for yy: 89360,000y2=178\frac{89 \cdot 360,000}{y^2} = 178 y2=89360,000178y^2 = \frac{89 \cdot 360,000}{178} y2=180,000y^2 = 180,000 y=180,000=424.26fty = \sqrt{180,000} = 424.26 \, \text{ft}

  3. Substitute y=424.26y = 424.26 back into the area equation to find xx: x=360,000424.26848.53ftx = \frac{360,000}{424.26} \approx 848.53 \, \text{ft}

Step 6: Calculate the Total Cost

Now that we have the dimensions:

  • x=848.53x = 848.53 ft
  • y=424.26y = 424.26 ft

The total fencing length needed is: Total Fencing=x+2y=848.53+2424.26=1697.05ft\text{Total Fencing} = x + 2y = 848.53 + 2 \cdot 424.26 = 1697.05 \, \text{ft}

The cost is: C=891697.05151,438.45C = 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 yy 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.

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Math Problem Analysis

Mathematical Concepts

Optimization
Differentiation
Cost Minimization
Area Calculation

Formulas

Area = x * y
Cost = 89 * (x + 2y)
Differentiation of Cost with respect to y

Theorems

Optimization in Calculus
Critical Points for Minimization

Suitable Grade Level

Grades 10-12