a farmer has 100 meters of fencing from which to build a rectangle chicken run. he intends using two adjacent walls for two sides of the rectangular enclosure
2 answers
x+y=50
No, the two sides add up to 100 meters
x + y = 100
You do not say what the question is.
If it is to maximize the area that you can get with that length of fence then:
A = x y
A = x (100-x)
without calculus
x^2 - 100 x = -A
x*2 - 100 x + 2500 = -A+2500
(x - 50)^2 = -1 (A-2500)
vertex of parabola at x = 50, Area = 2500
with calculus
A = x (100-x)
A = 100 x - x^2
dA/dx = 100 - 2 x
for max or min dA/dx = 0
2 x = 100
x = 50 again
x + y = 100
You do not say what the question is.
If it is to maximize the area that you can get with that length of fence then:
A = x y
A = x (100-x)
without calculus
x^2 - 100 x = -A
x*2 - 100 x + 2500 = -A+2500
(x - 50)^2 = -1 (A-2500)
vertex of parabola at x = 50, Area = 2500
with calculus
A = x (100-x)
A = 100 x - x^2
dA/dx = 100 - 2 x
for max or min dA/dx = 0
2 x = 100
x = 50 again
1 answer