A farmer decides to make three identical pens with 64 feet of fence. The pens will be next to each other sharing a fence and will be up against the barn. The bard side needs no fence. What dimensions for the total enclosure (rectangle including all pens) will make the area large as possible?
To find the dimensions of the total enclosure that will make the area as large as possible, we need to maximize the area of the enclosure.
Let's assume the length of each pen is x and the width is y.
Since three identical pens will be next to each other, the total length of the three pens will be 3x. We know that the total fence length available is 64 feet, and since one side is against the barn and doesn't need a fence, we can write the equation:
3x + y = 64
To find the maximum area, we need to express the area in terms of one variable. The area of each pen is x * y, and since there are three identical pens, the total area of the pens is 3xy.
The width of the total enclosure (rectangle including all pens) will be the same as the width of the pens, which is y.
The length of the total enclosure will be the total length of the three pens plus the width of one pen, which is 3x + y.
Therefore, the area of the total enclosure is (3x + y) * y = 3xy + y^2.
To maximize the area, we need to find the maximum value of this quadratic equation.
We can rewrite the equation 3x + y = 64 as y = 64 - 3x.
Substituting this equation into the area equation, we get:
Area = 3x(64 - 3x) + (64 - 3x)^2
Expanding and simplifying, we get:
Area = 192x - 9x^2 + 4096 - 384x + 9x^2
Combine like terms:
Area = -192x + 4096
The area (A) in terms of x is a linear function with a negative slope. The maximum area will occur at the x-coordinate of the vertex of this linear function.
The x-coordinate of the vertex can be found using the formula: x = -b / (2a), where a = -192 and b = 0.
x = -0 / (2 * -192)
x = 0
Therefore, the maximum area occurs when x = 0.
Since x represents the length of each pen, a length of 0 feet is not possible or practical.
Therefore, there is no solution to this problem under the given conditions, as it is not possible to create identical pens with non-zero lengths that maximize the area of the total enclosure with 64 feet of fence.