SOLUTION: A farmer has 70 meters of fencing and would like to use the fencing to create a rectangular garden where one of the sides of the garden is against the side of a barn.
Let l repre
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Question 1126355: A farmer has 70 meters of fencing and would like to use the fencing to create a rectangular garden where one of the sides of the garden is against the side of a barn.
Let l represent the varying length of the rectangular garden (in meters) and let A
represent the area of the rectangular garden (in square meters).
A)Write a formula that expresses A in terms of l.
B)What is the maximum area of the garden?
C)What is the length and width of the garden configuration that produces the maximum area?
D)What if the farmer instead had 300 meters of fencing to create the garden. What is the length and width of the garden configuration that produces the maximum area?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A farmer has 70 meters of fencing and would like to use the fencing to create a rectangular garden where one of the sides of the garden is against the side of a barn.
Let L represent the varying length of the rectangular garden (in meters) and
let A represent the area of the rectangular garden (in square meters).
:
let w = the width of the garden
then three fenced sides
L + 2w = 70
2w = -L + 70
w = -.5L + 35
:
A)Write a formula that expresses A in terms of L.
A(L) = L(-.5L+35)
A(L) = -.5L^2 + 35L
:
B)What is the maximum area of the garden?
find the axis of symmetry of equation: A = -.5L^2 + 35L
L =
L = +35 meters
A = -.5(35^2) + 35(35)
A = 612.5 sq/m
:
C)What is the length and width of the garden configuration that produces the maximum area?
L = 35 m is the length
w = -.5(35) + 35
w = 17.5 m is the width
:
D)What if the farmer instead had 300 meters of fencing to create the garden. What is the length and width of the garden configuration that produces the maximum area?
L + 2w = 300
2w = -L + 300
w = -.5L + 150
:
A)Write a formula that expresses A in terms of l.
A(L) = L(-.5L+150)
A(L) = -.5L^2 + 150L
:
B)What is the maximum area of the garden?
find the axis of symmetry of equation: A = -.5L^2 + 150L
L =
L = 150 meters is the length
then
A = -.5(150^2) + 35(150)
A = 11,250 sq/m
:
w = -.5(150) + 150
w = 75 m is the width
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