Math Problem Statement

A farmer wants to build a rectangular pen along a straight river. She wants to divide the pen into 5 equal rectangular pieces as shown in the picture. She will not fence one of the long sides along the river. (a) What is the largest area she can enclose with 3,000 feet of fencing? (b) What are the length and width of the individual pens that will give the largest area?

Solution

This problem involves maximizing the area of a rectangular pen that is divided into five equal pieces, without fencing along one side of the rectangle (since the river acts as a natural boundary).

Let’s break this down and solve it step by step.

Step 1: Define the variables

  • Let LL be the length of the pen (the side parallel to the river).
  • Let WW be the width of the pen (the side perpendicular to the river).

Since the pen is divided into 5 smaller rectangular sections, there are 4 vertical divisions required along the width.

Step 2: Set up the equation for the fencing

The total length of fencing available is 3,000 feet. The fencing is used for:

  • One side of length LL (opposite to the river),
  • Two sides of width WW (on the left and right),
  • Four internal vertical dividers along the width.

The total length of the fencing can therefore be expressed as: Total Fencing=L+6W=3000feet\text{Total Fencing} = L + 6W = 3000 \, \text{feet}

This gives us the constraint: L+6W=3000L + 6W = 3000

Step 3: Express LL in terms of WW

We solve for LL in terms of WW: L=30006WL = 3000 - 6W

Step 4: Set up the area function

The area of the entire pen is given by: A=L×WA = L \times W Substituting L=30006WL = 3000 - 6W into the area formula: A(W)=(30006W)×W=3000W6W2A(W) = (3000 - 6W) \times W = 3000W - 6W^2

Step 5: Maximize the area function

To find the value of WW that maximizes the area, we take the derivative of A(W)A(W) with respect to WW and set it equal to zero.

The derivative of A(W)A(W) is: dAdW=300012W\frac{dA}{dW} = 3000 - 12W Setting this equal to zero: 300012W=03000 - 12W = 0 W=250feetW = 250 \, \text{feet}

Step 6: Find LL

Now, substitute W=250W = 250 feet into the equation L=30006WL = 3000 - 6W: L=30006(250)=30001500=1500feetL = 3000 - 6(250) = 3000 - 1500 = 1500 \, \text{feet}

Step 7: Calculate the maximum area

The maximum area is: A=L×W=1500×250=375,000square feetA = L \times W = 1500 \times 250 = 375,000 \, \text{square feet}

Final Answer:

(a) The largest area the farmer can enclose is 375,000 square feet.

(b) The dimensions of the individual pens that will give the largest area are:

  • Length of each pen: L=1500feetL = 1500 \, \text{feet}
  • Width of each pen: W=250feetW = 250 \, \text{feet}

Questions:

  1. Why does maximizing the width result in maximizing the overall area?
  2. How would the solution change if more than 3,000 feet of fencing were available?
  3. Could the farmer enclose a larger area by changing the number of divisions (e.g., 4 pens instead of 5)?
  4. What happens to the area if the fencing along the river was required?
  5. How would the solution change if the river wasn't straight?

Tip:

Always express one variable in terms of the other when dealing with constrained optimization problems to reduce the number of variables in the equation.

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Math Problem Analysis

Mathematical Concepts

Optimization
Algebra
Differentiation
Quadratic Functions

Formulas

Fencing constraint: L + 6W = 3000
Area function: A = L * W = (3000 - 6W) * W = 3000W - 6W^2
Derivative: dA/dW = 3000 - 12W

Theorems

Maximization using first derivative
Quadratic Optimization

Suitable Grade Level

Grades 11-12 (Advanced High School)