A farmer wants to fence in an area of 15000 m² in a rectangular field and then divide it into half with a fence parallel to one sides of the rectangle. How can he do this so as to minimize the cost of the fence?
Are you assuming 150 by 100? If so, the amount of fence would be 3(100)+2(150). Compare that to other possibilities, e.g., 300 by 50.
Let the long side by y m
let the shorter side be x m
So the area is xy and xy = 15000, -----> y = 15000/x
For the cost to be minimum , the total amount of fencing has to be minimum
F = 2y + 3x
= 2(15000/x) + 3x
dF/dx = -30000/x^2 + 3
= 0 for a minimum of F
3 = 30000/x^2
x^2 = 10000
x = 100 , then y = 15000/x = 150
State your conclusion
Well, if the farmer wants to minimize the cost of the fence, he should think outside the box! Or in this case, outside the rectangle. Let's get creative, shall we?
Instead of using just any random rectangle, why not go for a square? A square is a rectangle, after all, just with a whole lot more equal sides!
So, the farmer can divide his 15,000 m² rectangular field into two equal halves by making a square in the middle. This way, he only needs to fence three sides of the square since he can utilize one side of the field as a natural boundary.
By doing this, he can minimize the cost of the fence, since using less fencing material means less expense. Plus, the square shape adds a touch of elegance to the overall design, making the field the talk of the town!
Remember, when it comes to minimizing costs, thinking outside the rectangle, or rather inside the square, can really make a difference!
To minimize the cost of the fence, the farmer should choose a rectangle with the smallest possible perimeter.
Step 1: Find the dimensions of the rectangle.
Let's assume the length of the rectangular field is L and the width is W.
Since the area of the rectangular field is given as 15000 m², we have the equation: L * W = 15000.
Step 2: Express one of the dimensions in terms of the other.
Solving the equation from Step 1 for L, we get: L = 15000 / W.
Step 3: Express the perimeter in terms of one variable.
The perimeter of the rectangle is: P = 2L + 2W.
Step 4: Substitute the expression for L from Step 2 into the equation from Step 3.
Substituting the expression for L in terms of W, we have: P = 2(15000 / W) + 2W.
Step 5: Get the derivative of P with respect to W.
To find the minimum value of P, we differentiate it with respect to W: dP/dW = -30000 / W^2 + 2.
Step 6: Set the derivative equal to zero and solve for W.
Setting dP/dW = 0, we have: -30000 / W^2 + 2 = 0.
Solving this equation, we get: 30000 / W^2 = 2.
This simplifies to: 15000 / W^2 = 1.
Step 7: Solve for W.
Simplifying further, we have W^2 = 15000.
Taking the square root of both sides, we find: W = √15000 ≈ 122.47 meters.
Step 8: Find the corresponding L value.
Using the equation L = 15000 / W, we have: L = 15000 / 122.47 ≈ 122.47 meters.
Step 9: Calculate the perimeter.
Using the formula P = 2L + 2W, we calculate the perimeter: P = 2(122.47) + 2(122.47) = 244.94 + 244.94 ≈ 489.89 meters.
Therefore, to minimize the cost of the fence, the farmer should choose a rectangle with dimensions approximately 122.47 meters by 122.47 meters, resulting in a total perimeter of approximately 489.89 meters.
To minimize the cost of the fence, the farmer should know that reducing the length of the fence will decrease the cost. Let's assume the length of the rectangular field is x meters and the width is y meters. To divide the field in half, he should place the fence parallel to one side, let's say the length.
Given that the area of the rectangular field is 15000 m², we have the equation xy = 15000.
To divide the field into half, the two halves will have equal areas, so each half will have an area of 7500 m².
Since the fence is parallel to the length, it splits the rectangular field into two equal halves with dimensions x meters by y/2 meters.
Therefore, the area of each half is (x)(y/2). From the equation above, we know that (x)(y) = 15000, so we can substitute these values into the area equation:
(x)(y/2) = 7500
Dividing both sides by (x/2), we get:
y/2 = 7500 / (x/2)
y/2 = 7500 / x
Solving for y, we have:
y = 7500 * 2 / x
y = 15000 / x
Now, we have the equation for y in terms of x.
To minimize the cost of the fence, we need to find the minimum value of the cost function, which is the sum of the lengths of all sides of the rectangular field.
The cost function can be expressed as: C = 2x + y
Substituting the value of y, we have:
C = 2x + (15000 / x)
Differentiating the cost function with respect to x, we have:
dC/dx = 2 - 15000 / x^2
To minimize the cost, we set the derivative equal to zero:
2 - 15000 / x^2 = 0
Simplifying the equation, we get:
2x^2 - 15000 = 0
Rearranging the equation, we have:
2x^2 = 15000
Dividing both sides by 2, we get:
x^2 = 7500
Taking the square root of both sides, we have:
x = sqrt(7500)
Now that we have the value of x, we can substitute it back into the equation for y to find its value:
y = 15000 / x
y = 15000 / sqrt(7500)
Simplifying further, we get:
y = sqrt(7500)
Therefore, to minimize the cost of the fence, the farmer should build a rectangular field with dimensions approximately x = sqrt(7500) meters by y = sqrt(7500) meters, and then place the fence parallel to one side of length x.