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).
:
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 = {{{(-35)/(2*-.5)}}}
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 = {{{(-150)/(2*-.5)}}}
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