A homeowner has forty feet of fencing to enclose three sides of a rectangular patio. The fourth side of the patio is a side of the house, and requires no fencing.
1. Let x represent the side of the rectangle perpendicular to the house (this helps avoid having ½ in the equation) and write an expression for the area as a function of x.
2. Give the domain by examining the equation from number 1.
3. Find the zeros of the equation.
4. Plot the zeros on a graph and find the x-coordinate of the vertex. Find the y-coordinate and make a sketch of the graph on the domain of x.
5. What should be the dimensions of the enclosure in order to maximize the area? What is the maximum area?
1. Ok with your defininition of x
let the other side (parallel to house) by y
so y + 2x = 40
y = 40-2x
Area = xy = x(40-2x) or
Area = 40x - 2x^2
2. clearly 0<x<20 or else the bracket (40-2) becomes negative.
3.
40x - 2x^2 = 0
2x(20-x) = 0
x = 0 or x = 20
4. you should get a vertex of (10,200)
x=10 , y = 20
The management of the UNICO department store has decided to enclose a 940 ft2 area outside the building for displaying potted plants and flowers. One side will be formed by the external wall of the store, two sides will be constructed of pine boards, and the fourth side will be made of galvanized steel fencing. If the pine board fencing costs $5/running foot and the steel fencing costs $3/running foot, determine the dimensions of the enclosure that can be erected at minimum cost
1. To find the expression for the area as a function of x, we need to find the perimeter of the rectangular patio using the given information.
Let's start by assigning variables to the sides of the rectangle:
Length of the patio = x
Width of the patio = y (which is also the side of the rectangle that needs fencing)
The perimeter of the rectangle is given by:
Perimeter = 2*(Length + Width)
Since we know the length of the patio (x) and three sides need fencing (perimeter minus the side adjacent to the house), we can write the expression for the area as follows:
Area = x * (40 - 2x)
2. The domain can be determined by considering the restrictions imposed by the problem. In this case, x represents the side of the rectangle perpendicular to the house. The length of a physical object cannot be negative, so the domain of x should be x ≥ 0.
3. To find the zeros of the equation, we set the expression for the area equal to zero and solve for x:
x * (40 - 2x) = 0
The equation will be zero if either x = 0 or (40 - 2x) = 0.
Solving the equation (40 - 2x) = 0, we get:
40 - 2x = 0
2x = 40
x = 40/2
x = 20
So, the zeros of the equation are x = 0 and x = 20.
4. To plot the zeros on a graph, we will have x values of 0 and 20. To find the x-coordinate of the vertex, we take the average of the zeros. The x-coordinate of the vertex is (0 + 20)/2 = 10.
To find the y-coordinate of the vertex, substitute the x-coordinate into the expression for the area:
Area = x * (40 - 2x)
Area = 10 * (40 - 2*10)
Area = 10 * (40 - 20)
Area = 10 * 20
Area = 200
So, the coordinates of the vertex are (10, 200).
Now, let's make a sketch of the graph on the domain of x:
The graph will be a quadratic function with a downward opening parabola. The vertex is at (10, 200), and the x-axis intercepts are at (0, 0) and (20, 0).
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5. To maximize the area, we need to find the dimensions of the enclosure. Since we know the length of the patio is x, the width (which requires fencing) can be calculated by subtracting twice the length from the total available fencing length:
Width = (40 - 2x)
To maximize the area, we need to find the maximum value of the area function. The maximum value occurs at the vertex of the parabola, which we found earlier to be (10, 200).
Therefore, to maximize the area, the dimensions of the enclosure should be:
Length = 10 feet
Width = (40 - 2*10) = 20 feet
The maximum area is 200 square feet.
1. To write an expression for the area as a function of x, we will use the formula for the area of a rectangle: A = length * width.
In this case, the length is x (perpendicular to the house) and the width is the remaining two sides that need fencing. Since we have 40 feet of fencing for three sides, we divide it by 3 to get the width of each side: 40 / 3 = 13.33 feet.
Therefore, the expression for the area, A, as a function of x is: A(x) = x * (40/3 - 2x).
2. The domain of the equation from number 1 represents the possible values that x can take. In this case, x represents the side of the rectangle perpendicular to the house.
Since x should be a positive value (you can't have a negative length), the domain is all positive real numbers. So the domain can be represented as: x > 0.
3. To find the zeros of the equation, we need to find the values of x that make the equation equal to zero. In other words, we need to solve the equation A(x) = 0.
A(x) = x * (40/3 - 2x) = 0
To solve this equation for x, we set each factor equal to zero:
x = 0 (this represents one of the zeros)
40/3 - 2x = 0
-2x = -40/3
x = 20/3 (this represents the other zero)
So the zeros of the equation are x = 0 and x = 20/3.
4. To plot the zeros on a graph, we can create an x-y coordinate system. The x-axis represents the side perpendicular to the house (x), and the y-axis represents the area (A).
Plotting the zeros:
- Zero 1: (0, 0)
- Zero 2: (20/3, 0)
To find the x-coordinate of the vertex, we can use the formula x = -b/(2a) from the vertex form of a quadratic equation. In this case, our equation is linear, not quadratic, but we can still use the formula.
For our equation A(x) = x * (40/3 - 2x), the coefficient of x^2 is 0 (since there is no x^2 term), so a = 0. The coefficient of x is -2, so b = -2.
Using x = -b/(2a), we find the x-coordinate of the vertex:
x = -(-2) / (2*0)
x = 2 / 0
x is undefined
Since x is undefined, we don't have a vertex for this linear equation.
5. To find the dimensions of the enclosure that maximize the area, we can use a method called optimization.
First, we know that the domain of x is x > 0, which means x can be any positive number. Next, we can analyze the behavior of the area function A(x) = x * (40/3 - 2x) within this domain.
Looking at the function, we can see that it is a quadratic equation in the form -2x^2 + (40/3)x. The coefficient of x^2 is negative, which means the graph of the function will be a downward-opening parabola.
Since the coefficient of x^2 is negative, the maximum area will occur at the vertex of the parabola. However, as previously determined, the linear equation does not have a vertex.
Therefore, in this case, since there is no vertex, the maximum area does not exist.