A Farmer With 750 Ft Of Fencing

A Farmer's Dilemma: Maximizing Area with 750 Feet of Fencing

A farmer finds himself facing a classic optimization problem: he has 750 feet of fencing and wants to create the largest possible rectangular enclosure for his livestock. This seemingly simple scenario opens the door to exploring key mathematical concepts, practical applications, and even some unexpected twists. Let's delve into how this farmer can maximize his land and explore variations on the problem.

Understanding the Basics: Perimeter and Area

The fundamental principle here is the relationship between the perimeter and area of a rectangle. The perimeter is the total length of the fencing, which is fixed at 750 feet. The area is the space enclosed within the fence, and this is what the farmer wants to maximize.

The formulas are:

• Perimeter (P) = 2l + 2w where 'l' is the length and 'w' is the width of the rectangle.
• Area (A) = l * w

Since the perimeter is fixed at 750 feet, we can rewrite the perimeter equation as:

750 = 2l + 2w

This equation allows us to express one variable in terms of the other. Let's solve for 'l':

l = (750 - 2w) / 2 = 375 - w

Now we can substitute this expression for 'l' into the area formula:

A = (375 - w) * w = 375w - w²

This equation represents a quadratic function. The graph of this function is a parabola that opens downwards, meaning there's a maximum point – the vertex – which represents the maximum area.

Finding the Maximum Area: Calculus Approach

To find the exact dimensions that maximize the area, we can use calculus. The derivative of the area function with respect to 'w' gives us the rate of change of the area:

dA/dw = 375 - 2w

Setting this derivative to zero and solving for 'w' will give us the width that maximizes the area:

375 - 2w = 0 2w = 375 w = 187.5 feet

Substituting this value back into the equation for 'l', we get:

l = 375 - 187.5 = 187.5 feet

Therefore, the dimensions that maximize the area are a square with sides of 187.5 feet. The maximum area is:

A = 187.5 * 187.5 = 35156.25 square feet

Finding the Maximum Area: Completing the Square

Alternatively, we can use the method of completing the square to find the vertex of the parabola representing the area function. The equation is:

A = -w² + 375w

Completing the square involves rewriting the equation in the form A = a(w - h)² + k, where (h, k) is the vertex. After completing the square, we get:

A = -(w² - 375w + (375/2)²) + (375/2)² A = -(w - 187.5)² + 35156.25

This clearly shows that the maximum area of 35156.25 square feet is achieved when w = 187.5 feet, confirming our previous result.

Considering Other Factors: A River Runs Through It

Let's add a twist to the problem. Suppose the farmer has a river running along one side of his property, eliminating the need for fencing on that side. This changes the perimeter equation and the optimization process.

Now, the perimeter equation becomes:

P = l + 2w = 750

Solving for 'l':

l = 750 - 2w

The area equation remains the same:

A = l * w = (750 - 2w) * w = 750w - 2w²

Using calculus, the derivative is:

dA/dw = 750 - 4w

Setting this to zero:

750 - 4w = 0 4w = 750 w = 187.5 feet

And solving for 'l':

l = 750 - 2(187.5) = 375 feet

This time, the optimal enclosure is a rectangle with dimensions 375 feet by 187.5 feet, resulting in a maximum area of:

A = 375 * 187.5 = 70312.5 square feet

Notice that using the river significantly increases the area the farmer can enclose.

Beyond Rectangles: Exploring Other Shapes

The problem assumes a rectangular enclosure. However, if the farmer is free to choose any shape, a circle would enclose the maximum area for a given perimeter. The formula for the circumference of a circle is:

C = 2πr

And the area is:

A = πr²

If C = 750, then r = 750 / (2π). Substituting this into the area formula, we get:

A = π * (750 / (2π))² ≈ 89522.4 square feet

This illustrates a crucial point: the shape of the enclosure dramatically affects the enclosed area. A circle, in theory, gives the largest area, although it might not be practical for livestock management.

Practical Considerations: Beyond the Math

While the mathematical solutions provide ideal dimensions, practical considerations come into play. The farmer might need to account for:

• Terrain: Uneven land might prevent the creation of a perfect rectangle or circle.
• Gate placement: Space must be left for a gate.
• Livestock behavior: The shape and size of the enclosure should suit the animals' needs.
• Accessibility: The farmer needs access to the enclosure for feeding, watering, and animal care.
• Existing infrastructure: The location of the enclosure might be constrained by existing structures or features.

Conclusion: Optimization in Action

The seemingly simple problem of a farmer with 750 feet of fencing highlights the power of mathematical optimization. Understanding the relationship between perimeter and area, using calculus or completing the square, allows us to determine the dimensions that maximize the area for a given perimeter. However, the real-world application requires considering practical constraints and adapting the theoretical solutions to create a functional and effective enclosure for the farmer's livestock. The exploration further extends to considering alternate shapes and how significantly these can impact the area enclosed. The farmer’s seemingly simple problem thus becomes a rich case study in applied mathematics, highlighting the interplay between theory and practice.