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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.
Step-by-step explanation:
The two equation will intersect each other at the point which will be the solution of the given two equations , and the given equations are ,
On subtracting the given equations we have,
Put this value in any equation , we have ,
Hence the lines will Intersect at ,