To make an enclosure for chickens, a rectangular area will be fenced next to a house. Only 3 sides will need to be fenced. There is 120 ft. of fencing material.
a. What quadratic function represents the area of the rectangular enclosure, where x is the distance from the house?
b. what dimensions will maximize the area of the enclosure?
Let the length parallel to the house be y
let each of the other two equal sides be x
2x + y = 120
y = 120 - 2x
Area = xy
= x(120 - 2x) or -2x^2 + 120x
b) you want the vertex.
x of the vertex is -120/(-4) = 30
when x = 30
y = 60
Maximum area = xy = 1800 ft^2
a. -2x²+120x
b. max area= 1800ft²
a. Well, let's break it down. We have a rectangular enclosure with one side being the house, and the other three sides being fenced. The house side doesn't need to be fenced, so we only need to worry about the other three sides. Since the sides are equal in length, we can let each side be represented by x. Therefore, the total perimeter of the enclosure is 3x. But we're given that we have 120 ft of fencing material, so we can set up the equation 3x = 120. Simplifying, we get x = 40.
Now, the area of the rectangular enclosure is given by the function A(x) = x * (120 - 2x). Simplifying further, we get A(x) = 120x - 2x^2. And there you have it, the quadratic function that represents the area of the rectangular enclosure is A(x) = 120x - 2x^2.
b. To maximize the area of the enclosure, we need to find the vertex of the quadratic function. The x-coordinate of the vertex can be found using the formula x = -b / (2a), where a = -2 and b = 120. Plugging in the values, we get x = -120 / (2 * -2) = 30.
So, the x-coordinate of the vertex is 30. Since we set x as the distance from the house, this means that the two sides of the enclosure that need to be fenced will each have a length of 30 ft. Therefore, the dimensions that will maximize the area of the enclosure are 30 ft x 30 ft.
To find the quadratic function that represents the area of the rectangular enclosure, we need to determine the dimensions of the enclosure in terms of x.
Let's assume the side adjacent to the house has a length of x ft. The opposite side will also have a length of x ft to make the rectangle symmetrical. The remaining side will be the width of the enclosure, which we can represent as y.
Now, let's calculate the length of the fence used. We have two sides with a length of x ft each, and one side with a length of y ft. So, the total length of the fence used is given by the equation:
2x + y = 120
To find the quadratic function representing the area, we need to express the area of the rectangle in terms of x.
The area of a rectangle is given by the formula: A = length × width.
In this case, the length is x ft and the width is y ft. Therefore, the area of the rectangle is:
A = x × y
Now we have two equations:
2x + y = 120 (equation for the perimeter)
A = x × y (equation for the area)
To find the quadratic function representing the area, we can solve the first equation for y and substitute it into the area equation:
2x + y = 120
y = 120 - 2x
Substituting this value of y into the area equation:
A = x × (120 - 2x)
Simplifying:
A = 120x - 2x^2
Thus, the quadratic function representing the area of the rectangular enclosure is A = 120x - 2x^2.
To find the dimensions that maximize the area of the enclosure, we need to find the value of x that results in the maximum value for the area function.
For a quadratic function, the maximum or minimum value occurs at the vertex of the parabola. The x-coordinate of the vertex is given by the formula x = -b/2a, where a and b are the coefficients of the quadratic function.
In this case, a = -2 and b = 120, so the x-coordinate of the vertex is:
x = -120 / (2(-2))
x = -120 / -4
x = 30
Therefore, the dimensions that maximize the area of the enclosure are x = 30 ft and y = 120 - 2(30) = 60 ft.
Hence, the dimensions that maximize the area of the enclosure are 30 ft by 60 ft.