A farmer has 25 yards of fencing to make a pig pen. He is going to use the side of the barn as one of the sides of the fence, so he only needs to fence 3 sides. What should be the dimensions of the fence in order to maximize the area?
let the side parallel to the barn be y yards
then the other two sides are x yards each
2x + y = 25
y = 25-2x
area = xy
= x(25-2x)
= 25x - 2x^2
which graphs as a downwards parabola, so the vertex has to be found.
quickest way:
the x of the vertex is -b/(2a) = -25/-4 = 6.25
then y = 25-12.5 = 12.5
the fields should be 12.5 long (along the barn) and 6.25 wide
To maximize the area of the pig pen, we need to find the dimensions that will give us the largest possible area. Let's assume the side parallel to the barn is x yards, and the other two sides are y yards each.
Since the perimeter of the pig pen is the sum of all three sides, we have:
x + y + y = 25
Simplifying this equation, we get:
x + 2y = 25
Now, we can express the area of the pig pen in terms of x and y. The area of a rectangle is given by the formula: Area = length * width. In this case, the length is x yards and the width is y yards. Thus, the area is:
Area = x * y
Our goal is to maximize the area, so we need to express the area in terms of a single variable. From the equation x + 2y = 25, we can solve for x:
x = 25 - 2y
Now, substitute this value of x back into the area equation:
Area = (25 - 2y) * y
Area = 25y - 2y^2
To find the dimensions that maximize the area, we need to find the vertex of the parabola representing the area equation. The x-coordinate of the vertex gives us the value of y that maximizes the area. We can find the vertex by using the formula: y = -b / (2a), where a = -2 and b = 25.
y = -25 / (2 * -2)
y = -25 / -4
y = 6.25
Therefore, the value of y that maximizes the area is y = 6.25 yards. We can now plug this value of y back into the equation x + 2y = 25 to find the value of x:
x + 2 * 6.25 = 25
x + 12.5 = 25
x = 25 - 12.5
x = 12.5
Hence, the dimensions that maximize the area of the pig pen are x = 12.5 yards and y = 6.25 yards.
To maximize the area of the pig pen, the dimensions should be in the form of a rectangle. Let's assume the dimensions of the rectangle as follows:
Length of the rectangle = L
Width of the rectangle = W
Since the farmer is using the barn as one side of the fence, the pig pen will have three sides. Hence, the perimeter of the pig pen can be calculated as:
Perimeter = L + W + L = 2L + W
According to the problem, the farmer has 25 yards of fencing, so the perimeter of the pig pen should be equal to 25 yards:
2L + W = 25
From the equation, we can express W in terms of L:
W = 25 - 2L
Now, let's consider the area of the rectangle:
Area = Length * Width = L * W
Substituting the expression for W:
Area = L * (25 - 2L) = 25L - 2L^2
To maximize the area, we need to find the value of L that makes the derivative of the Area function equal to zero.
Let's differentiate the Area function with respect to L:
d(Area)/dL = 25 - 4L
Setting the derivative equal to zero and solving for L:
25 - 4L = 0
4L = 25
L = 25/4
L = 6.25
So, the length of the rectangle should be 6.25 yards.
Substituting this value of L back into the equation for W:
W = 25 - 2L = 25 - 2(6.25) = 25 - 12.5 = 12.5
Therefore, the dimensions of the fence that would maximize the area is a rectangle of length 6.25 yards and width 12.5 yards.