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
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=50
To find the dimensions of the rectangular chicken run, we can use the given information that there are two adjacent walls already in place. Let's assume the length of the rectangular enclosure will be along these two adjacent walls.
Let L be the length of the rectangular chicken run along the adjacent walls, and let W be the width of the rectangular chicken run perpendicular to the adjacent walls.
Now, let's consider the fence required to build this chicken run. The length of the rectangular chicken run (L) will require two fences, one for each of the adjacent walls. The width of the rectangular chicken run (W) will require two additional fences, one on each end.
So, the total fence required can be calculated as:
Total fence = 2L + 2W
Given that the farmer has 100 meters of fencing, we can write the equation:
2L + 2W = 100
We want to find the dimensions of the rectangular enclosure, so let's solve this equation for L in terms of W:
2L = 100 - 2W
L = (100 - 2W) / 2
L = 50 - W
Now, we can substitute this value of L in the equation to find the width:
2(50 - W) + 2W = 100
100 - 2W + 2W = 100
100 = 100
This equation doesn't have any variables, which means that the width (W) can have any value. Therefore, the width of the rectangular chicken run can be any value, as long as it is less than 50 meters, since it cannot exceed half the total length of the fencing.
However, there is no unique solution for the length and width of the rectangular chicken run given the information provided.
To find the dimensions of the rectangular chicken run, we need to consider that the farmer will use two adjacent walls, while the other two sides will be made from the fencing.
Let's assume the lengths of the two adjacent walls are L and W, and the lengths of the other two sides made from the fencing are L and W as well.
The perimeter of a rectangle is calculated by adding up the lengths of all four sides, which in this case is equal to 100 meters:
Perimeter = 2L + 2W
Given that the perimeter is 100 meters, we can rewrite the equation as:
2L + 2W = 100
Now we need to rearrange the equation and isolate one of the variables. Let's solve for L:
2L = 100 - 2W
L = (100 - 2W) / 2
L = 50 - W
Now we have an equation for L in terms of W. We can substitute this value of L into the perimeter equation:
2(50 - W) + 2W = 100
100 - 2W + 2W = 100
100 = 100
This equation is true, which means any value of W will satisfy it. In other words, the width W can be any value. However, we need to choose a value for W that makes sense in the context of a chicken run.
Let's assume the farmer wants a reasonably sized chicken run, so W should be greater than zero but less than 50 (since the length L cannot be negative or greater than 50). Let's say we choose a width of 40 meters:
W = 40
Now we can find the length L by substituting this value back into the equation for L:
L = 50 - W
L = 50 - 40
L = 10
Therefore, if the farmer uses two adjacent walls for two sides of the rectangular enclosure and has 100 meters of fencing, the dimensions of the chicken run would be 10 meters by 40 meters.