Math Problem Statement

A rectangular field is to be fenced along two sides and two ends. It is also required to have three additional cross fences from side to side to form four plots. Find the length of the field to the nearest tenth that will give the entire field a maximum area if 414 meters of fencing is available.

Solution

Let's break down the problem to solve it step by step.

Given:

  • The total fencing available is 414 meters.
  • The field is rectangular, with two sides of width ww and two ends of length LL.
  • Additionally, there are three cross fences dividing the field into four plots, each of width ww.

Fencing Equation:

The total amount of fencing used includes:

  • The two lengths of the field: 2L2L.
  • The two widths of the field: 2w2w.
  • The three cross fences, which together add up to 3w3w since each cross fence is also of width ww.

Thus, the total fencing constraint can be expressed as: 2L+5w=4142L + 5w = 414

Area of the Field:

The area AA of the field is given by the formula: A=L×wA = L \times w

Objective:

We need to maximize the area while adhering to the fencing constraint. To do this, we first solve for LL in terms of ww from the fencing equation.

2L=4145w2L = 414 - 5w L=4145w2L = \frac{414 - 5w}{2}

Now, substitute this expression for LL into the area formula: A=w×4145w2A = w \times \frac{414 - 5w}{2} A=w(4145w)2A = \frac{w(414 - 5w)}{2} A=414w5w22A = \frac{414w - 5w^2}{2} A=207w52w2A = 207w - \frac{5}{2}w^2

Maximizing the Area:

To maximize the area, we take the derivative of AA with respect to ww, and set it equal to zero to find the critical points. dAdw=2075w=0\frac{dA}{dw} = 207 - 5w = 0 5w=2075w = 207 w=2075=41.4w = \frac{207}{5} = 41.4

Finding the Corresponding Length LL:

Now, substitute w=41.4w = 41.4 back into the equation for LL: L=4145(41.4)2L = \frac{414 - 5(41.4)}{2} L=4142072L = \frac{414 - 207}{2} L=2072=103.5L = \frac{207}{2} = 103.5

Thus, the length of the field is 103.5 meters.

Final Answer:

The length of the field that maximizes the area, with 414 meters of fencing, is 103.5 meters.

Would you like further details or have any questions?


Here are five questions to explore this problem further:

  1. How do you determine when a function reaches its maximum or minimum value?
  2. Why do we take the derivative to find the maximum area?
  3. What would happen if more or fewer cross fences were added?
  4. How does the area change if we increase the total amount of fencing?
  5. Could we use another optimization method to solve this problem?

Tip: In optimization problems, always check if you're maximizing or minimizing the function, and apply constraints accordingly.

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Math Problem Analysis

Mathematical Concepts

Optimization
Quadratic Equations
Derivatives

Formulas

Total fencing constraint: 2L + 5w = 414
Area of the field: A = L * w
Derivative of area: dA/dw = 207 - 5w

Theorems

Maximizing area using derivatives
Solving quadratic equations

Suitable Grade Level

Grade 11-12