A sample of 12 measurements has a mean of 38 and a sample standard deviation of 4.25. Suppose that the sample is enlarged to 14
1 answer:
You might be interested in
Given:
l = length of the rectangle
w = width of the rectangle
P = 4 ft, constant perimeter
Because the given perimeter is constant,
2(w + l) = 4
w + l = 2
w = 2 - l (1)
Part A.
The area is
A = w*l
= (2 - l)*l
A = 2l - l²
This is a quadratic function or a parabola.
Part B.
Write the parabola in standard form.
A = -[l² - 2l]
= -[ (l -1)² - 1]
= -(l -1)² + 1
This is a parabola with vertex at (1, 1). Because the leading coefficient is negative the curve is downward, as shown below.
The maximum value occurs at the vertex, so the maximum value of A = 1.
From equation (1), obtain
w = 2 - l = 2 - 1 = 1.
The maximum value of the area occurs when w=1 and l=1 (a square).
Answer:
The area is maximum when l=1 and w=1.
The geometric argument is based on the vertex of the parabola denoting maximum area.
l = length of the rectangle
w = width of the rectangle
P = 4 ft, constant perimeter
Because the given perimeter is constant,
2(w + l) = 4
w + l = 2
w = 2 - l (1)
Part A.
The area is
A = w*l
= (2 - l)*l
A = 2l - l²
This is a quadratic function or a parabola.
Part B.
Write the parabola in standard form.
A = -[l² - 2l]
= -[ (l -1)² - 1]
= -(l -1)² + 1
This is a parabola with vertex at (1, 1). Because the leading coefficient is negative the curve is downward, as shown below.
The maximum value occurs at the vertex, so the maximum value of A = 1.
From equation (1), obtain
w = 2 - l = 2 - 1 = 1.
The maximum value of the area occurs when w=1 and l=1 (a square).
Answer:
The area is maximum when l=1 and w=1.
The geometric argument is based on the vertex of the parabola denoting maximum area.