Asked by Vertical line test
Find the dimensions of the largest rectangular garden that can be enclosed by 60m of fencing.
Answers
Answered by
Bosnian
Perimetet of rectangle:
P = 2 L + 2 W
In this case:
2 L + 2 W = 60
Divide both sides by 2
L + W = 30
Subtract W to both sides.
L = 30 - W
Area of a rectangle:
A = L • W = ( 30 - W ) • W =
30 W - W² = - W² + 30 W
Graph of this function is parabola opens downward.
Vertex if this parabola is the maximum.
W coordinate of this parabola vertex is:
W = - b / 2 a = - 30 / 2 • ( - 1 ) =
- 30 / - 2 = 15
W = 15 m
L = 30 - W = 30 - 15 = 15
L = 15 m
P = 2 L + 2 W
In this case:
2 L + 2 W = 60
Divide both sides by 2
L + W = 30
Subtract W to both sides.
L = 30 - W
Area of a rectangle:
A = L • W = ( 30 - W ) • W =
30 W - W² = - W² + 30 W
Graph of this function is parabola opens downward.
Vertex if this parabola is the maximum.
W coordinate of this parabola vertex is:
W = - b / 2 a = - 30 / 2 • ( - 1 ) =
- 30 / - 2 = 15
W = 15 m
L = 30 - W = 30 - 15 = 15
L = 15 m
Answered by
Bosnian
Remark:
L = length
W = width
A 15 m x 15 m square is the largest rectangle that can be enclosed with 60 m of a fence.
L = length
W = width
A 15 m x 15 m square is the largest rectangle that can be enclosed with 60 m of a fence.
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