A farmer has $2400 to spend to fence two rectangular pastures.The local contractor will build the fence at a cost of $6.25/m. What is the largest area that the farmer can have fenced for that price?
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The farmer can buy $2400 / $6.25/m = 384m of fence. 384m/2 = 192m for each of the 2 rectangular pastures. A square is the rectangle with the largest area for a given perimeter, so each square will have sides of 192m/4 = 48m. Each pasture will thus have area of 48m x 48m = 2304 sq.m, and so the total area for both pastures will be 2304 x 2 = 4608 sq. m.
To find the largest area that the farmer can have fenced for $2400, we need to determine the dimensions of the two rectangular pastures.
Let's assume that the length of the first pasture is L1 and the width is W1, and the length of the second pasture is L2 and the width is W2.
The cost of fencing the first pasture is given by the formula: Cost1 = Perimeter1 * Cost per meter
Cost1 = (2L1 + 2W1) * $6.25
Similarly, the cost of fencing the second pasture is given by: Cost2 = (2L2 + 2W2) * $6.25
We know that the total cost of fencing both pastures is $2400, so:
Cost1 + Cost2 = $2400
(2L1 + 2W1) * $6.25 + (2L2 + 2W2) * $6.25 = $2400
Simplifying the equation, we have:
12.5L1 + 12.5W1 + 12.5L2 + 12.5W2 = 2400
12.5(L1 + W1 + L2 + W2) = 2400
L1 + W1 + L2 + W2 = 192
We want to maximize the area, which is given by: Area = L1 * W1 + L2 * W2
To find the dimensions that maximize the area, we can use the fact that the sum of the sides of a rectangle is half of the perimeter, which is L + W = Perimeter / 2.
Therefore, we have:
L1 + W1 = (192 - (L2 + W2))/2
We can rearrange the equation to solve for W1:
W1 = (192 - (L2 + W2))/2 - L1
Substituting this into the area formula, we have:
Area = L1 * [(192 - (L2 + W2))/2 - L1] + L2 * W2
To find the maximum area, we need to take the derivative of the area formula with respect to L1 and set it equal to zero, and similarly for L2 and W2. However, finding the exact maximum of this function can be quite complicated.
Instead, we can use trial and error or numerical methods to find the dimensions that give the largest area within the given constraints.
To find the largest area that the farmer can have fenced for $2400, we need to maximize the area of the rectangular pastures.
Let's assume the length of one rectangular pasture is "x" meters and the width is "y" meters. Since there are two pastures, we have two rectangles.
The total cost of fencing can be calculated by finding the perimeter of both rectangles and multiplying it by the cost per meter, which is $6.25/m.
The perimeter (P) of a rectangle can be calculated using the formula: P = 2(x + y)
The total cost (C) of fencing both rectangular pastures can be calculated as: C = 2(x + y) * 6.25
Given that the total cost is $2400, we can set up the equation: 2(x + y) * 6.25 = 2400
Simplifying the equation: (x + y) = 2400 / (2 * 6.25)
(x + y) = 192
Now, we want to maximize the area (A) of the rectangle. The formula for the area of a rectangle is: A = x * y
Since we know the relationship (x + y) = 192, we can rewrite it as y = 192 - x
Substituting this value of y into the area formula, we get: A = x * (192 - x)
Now, we need to find the maximum value of A. To do this, we can differentiate A with respect to x and set it equal to zero.
dA/dx = 192 - 2x
192 - 2x = 0
2x = 192
x = 96
Now, substitute the value of x back into the equation to find y: y = 192 - 96 = 96
Therefore, the dimensions of each rectangular pasture should be 96m x 96m. Hence, the largest area that the farmer can have fenced for $2400 is 96m x 96m = 9216 square meters.