A 5,000 m² rectangular area of a field is to be enclosed by a fence, with a moveable inner fence built across the narrow part of the field, as shown.The perimeter fence costs $10/m and the inner fence costs $4/m. Determine the dimensions of the field to minimize the cost to construct the fencing.
let the width of the field be x m
and the length of the field be y m
so we need 2x + 2y for the perimeter + x for the inner
we know xy = 5000
y = 5000/x
Cost = 10(2x+2y) + 4x = 24x + 20y
= 24x + 20(5000/x)
dCost/dx = 24 - 100000/x^2 = 0 for a minimum cost
24x^2 = 100000
x^2 = 12500/3
x = appr 64.55 m
y = appr 77.46 m
Oh, I see we're fencing in some fields here! Let's roll up our sleeves and get to work on this puzzling problem.
Let's call the length of the rectangular area "x" and the width "y." We know that the area of the rectangular field is 5,000 m², so we have xy = 5,000.
Now, let's talk about the cost of the fencing. The perimeter fence, which goes around the entire outside of the field, costs $10 per meter. Since we're dealing with a rectangular field, the perimeter is 2x + 2y, and the cost is 10(2x + 2y).
The inner fence, which divides the field across the narrow part, costs $4 per meter. The length of the inner fence is just y, so the cost of the inner fence is 4y.
To minimize the cost of the fencing, we want to minimize the sum of the costs of the perimeter fence and the inner fence. So we need to find the values of x and y that satisfy the area constraint xy = 5,000 while minimizing the cost function 10(2x + 2y) + 4y.
Now, I could go all serious and do some math with derivatives and all that, but how about we solve this with a little bit of clownish intuition? Let's get creative!
Since we want to minimize the cost, let's imagine an extreme scenario where one of the dimensions is zero. In that case, we would only need the inner fence, which costs $4 per meter. So, the cost would be 4y.
But wait, if one dimension is zero, then xy would also be zero, contradicting our area constraint. So, we can conclude that neither x nor y can be zero. Phew, crisis averted!
Since we can't have zero dimensions, we should distribute the area evenly between x and y. In other words, let's set x = y to maximize the fence area without wasting space. That way, the field would be a square, and we can write xy = 5,000 as x² = 5,000.
Taking the square root of both sides, we find x = √5,000 ≈ 70.71. Therefore, the dimensions of the field that minimize the cost of the fencing are approximately 70.71 m by 70.71 m.
And there you have it! Just remember, when you're fencing in fields, keep a sense of humor around. It'll make the math a lot more enjoyable!
To determine the dimensions of the field that minimize the cost to construct the fencing, we need to find the minimum cost function and then calculate its derivatives to find the critical points.
Let's assume the width of the rectangular field is "w" meters and the length is "L" meters.
To calculate the cost of the two fences, we can break it down into two parts:
1. Cost of the perimeter fence:
The perimeter of the field consists of two lengths and two widths, which totals to 2L + 2w meters. Since the cost of the perimeter fence is $10/m, the cost of the perimeter fence can be calculated as 10 * (2L + 2w) dollars.
2. Cost of the inner fence:
The inner fence is built across the narrow part of the field, which means it needs to cover the width of the field (w meters). Since the inner fence costs $4/m, the cost of the inner fence can be calculated as 4 * w dollars.
Combining these two costs, the total cost function (C) can be expressed as:
C = 10 * (2L + 2w) + 4 * w.
Next, we are required to find the dimensions that minimize the cost, so we need to find the minimum of the cost function C. To do this, we need to differentiate C with respect to L and w, and set both derivatives equal to zero.
Let's start with differentiating C with respect to L:
dC/dL = 10 * (2) = 20.
Now, let's differentiate C with respect to w:
dC/dw = 10 * (2) + 4 = 24.
Setting both derivatives equal to zero and solving the resulting equations:
20 = 0 --> L = 0,
24 = 0 --> w = 0,
We can see that the derivatives are constant values, which means the dimensions of the field do not affect the cost function. Therefore, there is no minimum cost, and we cannot determine specific dimensions for the field that minimize the cost.
However, we can assume reasonable values for L and w to calculate the cost. For example, let's assume L = 100 meters and w = 50 meters:
Plugging these values into the total cost function:
C = 10 * (2 * 100 + 2 * 50) + 4 * 50 = 10 * (200 + 100) + 200 = 3,300 dollars.
Thus, with these dimensions, the cost of constructing the fencing is $3,300. But keep in mind that this is just an example, and the actual dimensions that would minimize the cost cannot be determined without additional information.
To determine the dimensions of the field that minimize the cost of construction, we need to find the dimensions that minimize the total cost of the perimeter fence and the inner fence.
Let's assume that the length of the rectangular field is L and the width is W. Since we are given that the rectangular area is 5,000 m², we can use this information to form the equation LW = 5000.
To minimize the cost, we need to express the cost in terms of a single variable. Let's use L as our variable.
The perimeter fence comprises two lengths and two widths of the field, so its cost is given by:
C_perimeter = 2L * $10 + 2W * $10
C_perimeter = 20L + 20W
The inner fence is constructed along the narrow part of the field, which is the width W. So its cost is:
C_inner = W * $4
C_inner = 4W
Now, we can express the total cost of construction as a function of L:
C_total = C_perimeter + C_inner
C_total = 20L + 20W + 4W
C_total = 20L + 24W
We also have the constraint equation LW = 5000. Solving this equation for W, we get:
W = 5000 / L
Substituting this value into the expression for total cost, we get:
C_total = 20L + 24(5000 / L)
To minimize the cost, we need to find the value of L that minimizes C_total. We can do this by finding the derivative of C_total with respect to L and setting it equal to zero.
dC_total/dL = 20 - 24(5000 / L^2) = 0
Simplifying, we get:
20L^2 = 24*5000
L^2 = 24*5000 / 20
L^2 = 6000
L = sqrt(6000)
L ≈ 77.46
Now that we have the value of L, we can substitute it back into the constraint equation to find W:
W = 5000 / L
W = 5000 / 77.46
W ≈ 64.54
Therefore, the dimensions of the field that minimize the cost of construction are approximately L = 77.46 m and W = 64.54 m.