Farmer KIKO wants to build a 150-square meter rectangular pigpen for his pigs. One wall of the
pen will be the existing barn for his pigs. The other three walls will be built from mesh (net)
fencing that cost $ 20/meter. Find the dimensions of the pigpen that will minimize Farmer KIKO’s
fencing cost.
minimizing the cost is the same as minimizing the fencing.
xy = 150
f = 2x+y = 2x + 150/x
df/dx = 2 - 150/x^2
df/dx=0 when x^2 = 75
So a pen of size √75 by 2√75 will use the least fencing.
Check: for a given perimeter, maximum area is when the fence is divided equally among lengths and widths. Since our total fence is 2√75 + 2√75 you can see that this is true.
To find the dimensions of the pigpen that will minimize Farmer KIKO's fencing cost, we need to minimize the perimeter of the pigpen while ensuring it has an area of 150 square meters.
Let's assume the length of the pigpen is x meters and the width is y meters.
The perimeter of the pigpen, P, can be calculated as follows:
P = 2x + y
Since one wall of the pen will be the existing barn, the perimeter equation can be simplified to:
P = x + 2y
Given that the area of the pigpen is 150 square meters, we have another equation:
xy = 150
To minimize the cost, we want to minimize the perimeter, or P. So we will solve for y in terms of x using the area equation, and substitute it into the perimeter equation:
xy = 150
y = 150/x
Substituting into the perimeter equation:
P = x + 2(150/x)
P = x + 300/x
To minimize P, we need to find the critical points of the equation by taking the derivative and setting it equal to 0:
dP/dx = 1 - 300/x^2
0 = 1 - 300/x^2
300/x^2 = 1
300 = x^2
x^2 = 300
x = √300
x ≈ 17.32 meters
Substituting the value of x back into the area equation to find y:
17.32y = 150
y ≈ 8.66 meters
Therefore, the dimensions of the pigpen that will minimize Farmer KIKO's fencing cost are approximately 17.32 meters in length and 8.66 meters in width.
To find the dimensions of the pigpen that will minimize Farmer KIKO's fencing cost, we can use optimization techniques.
Let's denote the dimensions of the rectangular pigpen as length (L) and width (W) in meters.
The area of the pigpen is given as 150 square meters, so we have:
L * W = 150
To minimize Farmer KIKO's fencing cost, we need to minimize the amount of fencing used. The total length of fencing required is the sum of lengths of the three sides of the pigpen, excluding the barn wall.
The total length of fencing required can be calculated as:
L + 2W
Since the cost of fencing is $20 per meter, we can express the cost of fencing as:
Cost = 20 * (L + 2W)
To minimize the cost, we can differentiate the cost function with respect to L and W separately and set the derivatives equal to zero.
∂Cost/∂L = 20 = 0
∂Cost/∂W = 40 = 0
Solving these equations will give us the optimal dimensions of the pigpen that minimize the fencing cost.
Differentiating the cost function with respect to L, we get:
∂Cost/∂L = 20
Differentiating the cost function with respect to W, we get:
∂Cost/∂W = 40
Since the derivatives are constants and cannot be equal to zero, we don't have any critical points to solve for.
Therefore, there are no values of L and W that will minimize the fencing cost. We can conclude that the cost will be minimized when L and W reach their minimum possible values, which would be a square pen with equal sides. In this case, the dimensions will be:
L = W = √150 ≈ 12.25 meters
This will give us a minimum cost of:
Cost = 20 * (12.25 + 2 * 12.25) = $860
So, Farmer KIKO should build a rectangular pigpen with dimensions of approximately 12.25 meters by 12.25 meters to minimize the fencing cost.
Why did the pig go to the gym?
Because he wanted to be a little bit more "sow"mall!
But let's get serious and calculate the dimensions of the pigpen.
Let's say the length of the pen is L and the width is W. Since one wall is already covered by the barn, we only need to fence the remaining three sides.
So the perimeter of the pigpen is:
P = 2L + W
And the area is given as 150 square meters:
A = L * W = 150
Now, let's focus on minimizing Farmer KIKO’s fencing cost. The cost of fencing is $20 per meter, so the total cost can be represented as:
Cost = P * 20 = (2L + W) * 20
We can solve this problem by substituting the value of W from the area equation.
A = L * W
150 = L * W
W = 150 / L
Substituting this in the cost equation, we get:
Cost = (2L + (150 / L)) * 20
To find the dimensions that minimize the cost, we need to find the minimum of this function. However, let me tell you a joke before I do the math:
Why did the pig become an architect?
Because he loved bringing "ham"ony to structures!
Alright, let's continue with the calculation.
To find the minimum cost, we can take the derivative of the cost function with respect to L and set it equal to zero:
dCost/dL = 2 - (150 / L^2) = 0
Simplifying this equation, we get:
2 = 150 / L^2
Cross-multiplying, we have:
2 * L^2 = 150
Dividing both sides by 2, we get:
L^2 = 75
Taking the square root of both sides, we find:
L = sqrt(75)
Now, substituting this value back into the equation for W:
W = 150 / L
W = 150 / sqrt(75)
So, the dimensions of the pigpen that will minimize Farmer KIKO’s fencing cost are approximately:
Length (L) ≈ sqrt(75)
Width (W) ≈ 150 / sqrt(75)
And that's how we find the optimal dimensions for the pigpen!