A clerk has 1/2 book of stamps on Monday. On Tuesday, he used 1/10 of the book of stamps. How much of the book of stamps is left
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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.