A farmer wants to enclose a rectangular field with 180m of fencing. The side of the barn will act as one side of the enclosure, leaving 3 sides to be covered. The function that describes the area inside is f(x)=x(180-2x) what are the dimensions of the closure with the largest area?
Thanks Damon! We got it now. We were stuck after line three so this was helpful.
Good, you are welcome
To find the dimensions of the enclosure with the largest area, we need to maximize the area function f(x) = x(180 - 2x).
To do this, we can find the critical points of the function by taking its derivative and setting it to zero:
f'(x) = 180 - 4x
Setting f'(x) = 0:
180 - 4x = 0
Solving for x:
4x = 180
x = 180/4
x = 45
So, x = 45 is a critical point of the function f(x). To determine if it is a maximum or minimum, we can take the second derivative:
f''(x) = -4
Since the second derivative is negative, it means that the critical point x = 45 is at a maximum.
Substituting x = 45 back into the original function f(x), we can find the maximum area:
f(x) = x(180 - 2x)
f(45) = 45(180 - 2(45))
f(45) = 45(180 - 90)
f(45) = 45(90)
f(45) = 4050
So, the enclosure with the largest area has dimensions of 45m and 90m.
180 x - 2 x^2 = A
2 x^2 -180 x = -A
x^2 - 90 x = - A/2
x^2 - 90 x + 45^2 = -A/2 + 45^2
(x-45)^2 = -(1/2) (A- 4050)
vertex at x = 45
so 45. 45 , 90
area = 4050