To find the coordinates of the vertex for the parabola defined by the quadratic function \( f(x) = -3x^2 + 5x - 4 \), we use the vertex formula for a parabola in the form \( ax^2 + bx + c \). The x-coordinate of the vertex is given by \( x = -\frac{b}{2a} \). Once we have the x-coordinate, we substitute it back into the function to find the y-coordinate.
For the function \( f(x) = -3x^2 + 18x - 2 \), we determine whether it has a minimum or maximum value by looking at the coefficient of \( x^2 \). If the coefficient is negative, the parabola opens downwards, indicating a maximum value. We find the maximum value by using the vertex formula as in the first question. The domain of any quadratic function is all real numbers, and the range can be determined from the vertex and the direction the parabola opens.
To find the dimensions of the rectangle that maximize the enclosed area with 1200 yards of fencing, we use the fact that the perimeter of the rectangle is 1200 yards. Let the length be \( l \) and the width be \( w \). The perimeter is given by \( 2l + 2w = 1200 \). We express one variable in terms of the other and use the area formula \( A = l \times w \) to find the maximum area using calculus or by completing the square.
For the quadratic function \( f(x) = -3x^2 + 5x - 4 \), we find the vertex using the formula for the x-coordinate of the vertex, \( x = -\frac{b}{2a} \). Substituting \( a = -3 \) and \( b = 5 \):
\[
x = -\frac{5}{2 \cdot -3} = \frac{5}{6} \approx 0.8333
\]
Next, we substitute \( x \) back into the function to find the y-coordinate:
\[
y = -3\left(\frac{5}{6}\right)^2 + 5\left(\frac{5}{6}\right) - 4 \approx -1.9167
\]
Thus, the coordinates of the vertex are:
\[
\boxed{\left(0.8333, -1.9167\right)}
\]
For the function \( f(x) = -3x^2 + 18x - 2 \), we again use the vertex formula. The coefficient of \( x^2 \) is negative, indicating that the function has a maximum value. The x-coordinate of the vertex is:
\[
x = -\frac{b}{2a} = -\frac{18}{2 \cdot -3} = 3
\]
Substituting \( x = 3 \) back into the function gives:
\[
y = -3(3)^2 + 18(3) - 2 = 25
\]
The domain of the function is all real numbers, and the range is:
\[
\text{Range} = (-\infty, 25]
\]
Thus, the maximum value and its location are:
\[
\boxed{(x = 3, y = 25)}
\]
Given a perimeter of 1200 yards, we set up the equation for the perimeter of a rectangle:
\[
2l + 2w = 1200
\]
This simplifies to:
\[
l + w = 600
\]
To maximize the area \( A = l \times w \), we express \( w \) in terms of \( l \):
\[
w = 600 - l
\]
Thus, the area becomes:
\[
A = l(600 - l) = 600l - l^2
\]
This is a quadratic function that opens downwards, and its maximum occurs at:
\[
l = \frac{600}{2} = 300
\]
Substituting back gives:
\[
w = 600 - 300 = 300
\]
The maximum area is:
\[
A = 300 \times 300 = 90000
\]
Thus, the dimensions of the rectangle that maximize the area are:
\[
\boxed{(l = 300, w = 300)}
\]
- Vertex of the parabola: \(\boxed{\left(0.8333, -1.9167\right)}\)
- Maximum value of the function: \(\boxed{(x = 3, y = 25)}\)
- Dimensions of the rectangle: \(\boxed{(l = 300, w = 300)}\)
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