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:
a. A = L * (90 - 2L)
b. To find the maximum area, we can take the derivative of the area function with respect to L and set it equal to zero:
dA/dL = (90 - 4L)
Setting this equal to zero:
90 - 4L = 0
4L = 90
L = 22.5
Substituting this value of L back into the area function:
A = 22.5 * (90 - 2(22.5))
A = 22.5 * (90 - 45)
A = 22.5 * 45
A = 1012.5 square meters
Therefore, the maximum area of the garden is 1012.5 square meters.
c. To find the length and width that produce the maximum area, we can use the value of L from above:
Length = L = 22.5 meters
Width = 90 - 2L = 90 - 2(22.5) = 45 meters
Therefore, the length and width of the garden configuration that produces the maximum area are 22.5 meters and 45 meters, respectively.
d. If the farmer had 260 meters of fencing, we can repeat the same process:
b. Maximum area = 260 * (90 - 2L)
c. Taking the derivative with respect to L and setting it equal to zero:
dA/dL = (90 - 2L)
Setting this equal to zero:
90 - 2L = 0
2L = 90
L = 45
Substituting this value of L back into the area function:
A = 45 * (90 - 2(45))
A = 45 * (90 - 90)
A = 45 * 0
A = 0 square meters
Therefore, with 260 meters of fencing, the maximum area is 0 square meters. This is because with the given constraints, it is not possible to create a rectangular garden.
a. To express A in terms of L, we need to determine the width of the garden first. Since one of the sides of the garden is against the side of a barn, the width will be constant. Let's call the width W.
Since a rectangular garden has two equal lengths (L) and two equal widths (W), the total amount of fencing required can be calculated as follows:
Fencing required = Length (L) + Length (L) + Width (W) = 2L + W
Given that the farmer has 90 meters of fencing, we can set up the equation:
2L + W = 90
Now, to find the area (A) of the rectangular garden, we multiply the length (L) by the width (W). Therefore:
A = L * W
b. To find the maximum area of the garden, we can solve the equation from part a for W in terms of L, and then substitute it into the area formula (A = L * W). However, since this approach might be complex, it would be easier to use a graphing calculator or computer software to plot the equation 2L + W = 90 and find the values of L and W that maximize the area.
c. To determine the length and width of the garden configuration that produces the maximum area, we would need to find the values of L and W that give the maximum value for A. This can be done by using the graphing calculator or computer software mentioned in part b.
d. If the farmer had 260 meters of fencing, we would follow a similar process as in parts a and b. We would set up the equation 2L + W = 260 and then use a graphing calculator or computer software to find the length (L) and width (W) that maximize the area (A). This would help us determine the values of L and W required for the garden with 260 meters of fencing.
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 :)
By the way the answer for max area is just cutting a square in half
w = L/2