A field is bound on one side by a river. A farmer wants to enclose the other three sides of the field with fence in order to create a rectangular plot of land for his cows. If the farmer has 400m of fence to work with, determine the maximum possible area of the field and the field's demensions.
(i'm on in grade 10, so I don't know calculus so don't use that)
10th grade? High time to learn calculus, on your own!
Anyway, we don' need no steeking calculus for this one.
If the side parallel to the river has length x, and the other side is y, then we have
x+2y = 400
The area a of the field is just xy, so
a = xy = (400-2y)y = 400y - 2y^2
This is just a parabola, opening downward, so it has a maximum value at the vertex, which occurs at y=400/4 = 100
so, if y=100, x=200 and the area is 20,000 m^2.
As is usual in this kind of problem, the maximum area is achieved when the fencing is divided equally among the lengths and widths. In this case 1 length of 200 and 2 widths of 100 each, making up the 400 total.
To determine the maximum possible area of the field, we need to find the dimensions that will maximize the area. Let's start by understanding the problem.
We know that the field is bound on one side by a river, meaning that only three sides of the field need to be fenced. This suggests that the river acts as a natural boundary and does not require additional fencing.
Let's suppose the length of the field is L and the width is W. Since the field is rectangular, we have two sides measuring L and two sides measuring W.
Considering that the farmer has 400m of fencing at their disposal, the total amount of fence used will be equal to the sum of all four sides of the rectangular field.
Therefore, we can express this as an equation:
2L + W = 400
To determine the maximum possible area, we need to eliminate one of the variables from the equation. For simplicity, let's solve the equation for W in terms of L:
W = 400 - 2L
Now we can express the area of the field in terms of a single variable:
Area = L * W
Substituting the value of W:
Area = L * (400 - 2L)
To find the maximum area, we need to find the value of L that maximizes this expression. As you mentioned, we won't use calculus, but rather an algebraic approach.
Let's simplify the equation:
Area = 400L - 2L^2
To find the maximum area, we can complete the square. However, in this case, since we're trying to understand how to find the answer rather than just finding the answer itself, let's consider using a table of values.
Let's tabulate the values of L and the corresponding areas with increments of, for example, 10 units:
L | Area
---------
0 | 0
10 | 3600
20 | 6400
30 | 8100
40 | 9600
50 | 10000
60 | 9600
70 | 8100
80 | 6400
90 | 3600
100| 0
By examining the table, we see that the maximum area of the field is 10,000 square meters, which occurs when L = 50. Therefore, the dimensions of the field would be a length of 50m and a width of 400 - 2(50) = 300m.
Hence, the maximum possible area of the field is 10,000 square meters, and the field's dimensions are 50m by 300m