length = L
width = w
fencing length = p = L + 2w = 90 so w = (90-L)/2
Area = a = L*w
a = L * (90-L)/2 = 45 L - .5 L*2
so
.5 L^2 - 45 L = -a
L^2 - 90 L = -2a ==== parabola opens down (sheds water)
complete square to find vertex
L^2 - 90 L + 45^2 = -2a + 45^2
(L-45)(L-45) = -2a + 45^2
L at vertex = 45
a at vertex = 45^2/2
etc :)
A farmer has 90 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
A=
b. What is the maximum area of the garden? (It may help to use a graphing calculator.)
c. What is the length and width of the garden configuration that produces the maximum area?
Length:
Width:
d. What if the farmer instead had 260 meters of fencing to create the garden. What is the length and width of the garden configuration that produces the maximum area?
Length:
Width:
2 answers
By the way the answer for max area is just cutting a square in half
w = L/2
w = L/2