A farmer wants to fence a rectangular area of 800,000 m² and divide it in half with a fence that is parallel to one of the sides of the rectangle, and twice as expensive as the fence on the outer sides. How can this be done in order to minimize the cost of the fence?
Hint: If all the fencing costs the same, what are we needing to minimize in this scenario?
I gave this a try but I keep getting stuck, i would really appreciate the help!
xy = 800000
If the outer fence costs $c/m, then the inner fence costs $2c/m
So, the total fence cost is
x(2*c + 2c) + y*c
so we want to minimize
z = 4x + 2*800000/x
dz/dx = 4 - 1600000/x^2 = 4(x^2-400000)/x^2
we want dz/dx=0 for a minimum, so
x^2 = 400000
x = 200√10
If all the fencing costs the same, then you want to minimize
z = 3x+2y
I interpret this rectangel to have 2 sides (lengths) and 3 sides (widths), with one of the widths cutting the rectangle is half
let the shorter sides be x each and each of the lengths equal y
so ...
xy = 800000 ---> y = 800000/x
You want to minimize the cost, so we need a "cost" equation
I also interpret that this one dividing fence cost twice as much as the others
so assume the cost of the outer fence is 1 unit each, then
cost = 2x(1) + 2y(1) + y(2)
= 2x + 4y
= 2x + 4(800000/x)
d(cost)/dx = 2 - 3200000/x^2 = 0 for a min cost
3,200,000/x^2 = 2
x^2 = 1,600,000
x = appr 1294.9 m
y = 632.46 m
check my arithmetic
if the cost is the same for all sides, then cost = 2x(1) + 3y(1)
proceed as before.
Well, let me try to bring some laughter into this mathematical dilemma!
Ahem, here's a little fence-related joke to lighten the mood: Why did the scarecrow win an award? Because he was outstanding in his field!
Now, let's tackle your question. We're looking to minimize the cost of the fence for the given scenario. Since the fence on the outer sides is half the price of the inner fence, we should aim to use more of the cheaper fence.
To do this, we can create a square-shaped enclosure by dividing the rectangle into two equal parts. This means each part will have an area of 400,000 m². By making the shape a square, we minimize the perimeter and consequently minimize the cost.
Think about it this way: a 1-meter by 400,000-meter rectangle would have a total cost of two long sides plus two short sides. On the other hand, a 200 meter by 200 meter square encloses the same area but with a smaller perimeter. And since the cost of the fence is directly related to the perimeter, using the square shape will give us the lowest cost.
I hope this solution brings a smile to your face, along with the joke earlier. If you have any more questions or need further assistance, feel free to ask!
To minimize the cost of the fence, we need to find the dimensions of the rectangular area that minimize the total length of the fence.
Let's assume the length of the rectangle is L and the width is W.
Since the fence that is parallel to one of the sides of the rectangle is twice as expensive, we can consider it as two separate fences. So, the cost of the outer sides fence will be C, and the cost of the inner fence (parallel to one of the sides) will be 2C.
We know that the area of the rectangle is 800,000 m², so we have the equation:
L * W = 800,000
Now, we need to express the length of the fence as a function of L and W.
The length of the outer sides of the rectangle is (2L + 2W), and the length of the inner fence is 2L.
So, the total length of the fence is:
F = (2L + 2W) + 2L
F = 4L + 2W
We want to minimize the total length of the fence F, so let's express F in terms of L only.
From the equation L * W = 800,000, we can express W in terms of L:
W = 800,000 / L
Substituting W in the equation for F:
F = 4L + 2(800,000 / L)
Now, we have F expressed in terms of L. To minimize F, we take the derivative of F with respect to L and set it equal to zero:
dF/dL = 4 - 1,600,000 / L^2 = 0
Solving this equation, we find:
L^2 = 1,600,000 / 4
L^2 = 400,000
L ≈ 632.5
Substituting this value of L back into the equation W = 800,000 / L, we find:
W ≈ 1265
Therefore, the dimensions of the rectangular area that minimize the cost of the fence are approximately:
Length (L) ≈ 632.5 m
Width (W) ≈ 1265 m
The total length of the fence is then:
F = 4L + 2W
F ≈ 4(632.5) + 2(1265)
F ≈ 2530 + 2530
F ≈ 5060 m
So, the minimum cost of the fence is achieved with a rectangular area of approximately 632.5 m by 1265 m, with a total length of fence being approximately 5060 meters.
To minimize the cost of the fence, we need to determine the dimensions of the rectangular area and the arrangement of the fences that will minimize the total length of fencing required.
Let's start by assuming the rectangular area has dimensions x meters by y meters. Since the area of the rectangle is given as 800,000 m², we have the equation:
x * y = 800,000 ...(Equation 1)
Next, we need to determine the arrangement of fences that minimizes the total length of fencing required.
The given condition states that a fence parallel to one of the sides of the rectangle will divide the area in half. Let's assume this dividing fence is parallel to the x-axis and splits the rectangle into two equal halves.
Now, if the dividing fence is parallel to the x-axis, it will divide the rectangle into two smaller rectangles, each with dimensions x meters by y/2 meters.
The total length of the fence required for the outer sides of the rectangle is:
L1 = 2 * (x + y) meters
The total length of the fence required for the dividing fence is:
L2 = 2 * x meters
According to the given condition, the cost of the dividing fence is twice as expensive as the cost of the outer sides. Therefore, the total cost of the fence is:
C = L1 + 2 * L2
Plugging in the expressions for L1 and L2, we get:
C = 2(x + y) + 4x
Simplifying further:
C = 6x + 2y
To minimize the cost C, we need to find the values of x and y that satisfy Equation 1 and minimize the cost function C.
To do this, we can use the given area equation:
x * y = 800,000
We can rewrite this equation as:
y = 800,000 / x
Substitute this expression for y in the cost equation C:
C = 6x + 2(800,000 / x)
To minimize C, we need to find the critical points by taking the derivative of C with respect to x and setting it to zero:
C' = 6 - 1,600,000 / x^2
Setting C' to zero:
6 - 1,600,000 / x^2 = 0
Solving for x:
1,600,000 / x^2 = 6
x^2 = 1,600,000 / 6
x^2 ≈ 266666.67
x ≈ √266666.67
x ≈ 516.4
Now, substitute the value of x back into the area equation to solve for y:
516.4 * y = 800,000
y ≈ 1,548.5
Thus, the dimensions of the rectangle that minimize the cost of the fence are approximately 516.4 meters by 1,548.5 meters.