One side of a rectangular pen is formed by a hedge. The other three sides are made using fencing.The length of the rectangle is x metre,120 metres of fencing is available.
a.show that the area of the rectangle is 1/2x*(120-x)m^2.
b.calculate the maximum posible area of the sheep pen.
length --- x
width ---- y
all you need is 1 length and 2 widths
x + 2y = 120
x = 120-2y
area = xy = y(120-2y) = 120y - 2y^2
d(area)/dy = 120 - 4y = 0 for a max of area
4y = 120
...
finish it up
a. To show that the area of the rectangle is 1/2x*(120-x)m^2, we need to express the area in terms of x.
The length of the rectangle is x meters, and the width can be calculated by subtracting twice the fence length (since there are two sides with the same length of fence) from the total available fencing (which is 120 meters):
Width = 120 - 2(x)
The area of a rectangle is given by length multiplied by width, so:
Area = x * (120 - 2x)
Simplifying further, we get:
Area = 120x - 2x^2
This can be rearranged to match the given form:
Area = -2x^2 + 120x
Multiplying both sides by -1 and factoring out -2, we get:
Area = -2(x^2 - 60x)
Expanding the equation, we get:
Area = -2(x(x - 60))
Now, we see that x(x - 60) represents the area of the rectangle.
Hence, the area of the rectangle is 1/2x*(120-x)m^2.
b. To calculate the maximum possible area of the sheep pen, we need to find the value of x that maximizes the area.
We can accomplish this by finding the vertex of the quadratic equation -2(x^2 - 60x).
The formula for finding the x-value of the vertex of a quadratic equation in the form ax^2 + bx + c is given by x = -b / (2a).
For our equation, a = -2, and b = -60.
Substituting these values into the formula:
x = -(-60) / (2*-2)
x = 60 / 4
x = 15
So, the value of x that maximizes the area is 15 meters.
Substituting this value back into the area equation, we can find the maximum possible area:
Area = -2(15^2 - 60*15)
Area = -2(225 - 900)
Area = -2(-675)
Area = 1350
Therefore, the maximum possible area of the sheep pen is 1350 square meters.
To show that the area of the rectangle is 1/2x * (120-x) square meters, we need to understand the structure of the pen.
a) Let's break down the rectangular pen into its components:
- One side of the rectangle is formed by a hedge, which we'll call length x meters.
- The other three sides are made using fencing, so each fence length will be (120 - x) meters. Since there are three fences, the total length of the three fence sides will be 3 * (120 - x) meters.
Now, let's calculate the area of the rectangle:
- The area of a rectangle is given by the formula: length * width.
- In this case, the length is x meters, and the width is 120 - x meters.
- Therefore, the area of the rectangle is x * (120 - x) square meters.
Now, let's simplify the expression for the area:
- Distributing x into (120 - x), we get: 120x - x^2.
- Rearranging the terms, we have: -x^2 + 120x.
- Now, let's factor out -1 from the expression: -1(x^2 - 120x).
- Factoring out x, we get: -1(x(x - 120)).
- Finally, multiplying by -1, we have: x(x - 120).
So, the area of the rectangle is 1/2x * (120 - x) square meters.
b) To find the maximum possible area of the sheep pen, we need to find the value of x that maximizes the area.
- The area of the rectangle is given by: A = 1/2x * (120 - x).
- To find the maximum area, we can differentiate this expression with respect to x and set it equal to zero.
- Differentiating A with respect to x, we get: dA/dx = (1/2)(120 - 2x).
- Setting dA/dx equal to zero, we have: (1/2)(120 - 2x) = 0.
- Simplifying, we get: 120 - 2x = 0.
- Solving for x, we find: x = 60.
To confirm that this point is a maximum, we can take the second derivative of A:
- The second derivative of A with respect to x is: d²A/dx² = -2.
- Since the second derivative is negative, this confirms that the point x = 60 is a maximum.
Therefore, the maximum possible area of the sheep pen is obtained when the length of the rectangle is 60 meters. By substituting x = 60 into the area formula, we find the maximum area to be 1/2 * 60 * (120 - 60), which equals 1/2 * 60 * 60 = 1800 square meters.