a farmer wants to make a rectangular pasture with 80000 squre meters. if the pasture lies along a river and he feneces the remaning three sides,what dimension shoud he use to minimize the amount fence needed
Taking the length as x, and the width as y,
Length of fence = x + y + x = 2x + y
Now, the area has been fixed.
=> xy = k = 80,000
=> y = k/x
Function to minimize = f(x) = Length
= 2x + y
= 2x + (k/x)
f'(x) = d(2x + (k/x))/dx
= 2 - k/(x^2)
Equating f'(x) with zero,
=> 2 - k/(x^2) = 0
=> 2 = k/x^2
=> x^2 = k/2 = 80000/2 = 40000
Hence, x = +200/-200
But length cannot be negative, so x = 200
=> x = 200, y = (80000/200) = 400
To minimize the amount of fence needed, the farmer should use the dimensions that would result in a rectangular pasture with the smallest perimeter.
Let's assume the length of the rectangular pasture along the river is "L" and the width perpendicular to the river is "W".
Given that the area of the rectangular pasture is 80,000 square meters, we can write the equation:
L × W = 80,000
Now, we need to minimize the amount of fencing required, which is equal to the perimeter of the rectangular pasture.
The perimeter, P, can be calculated using the formula:
P = 2L + W
To find the smallest perimeter, we can solve for W or L.
Let's solve for W in terms of L using the equation L × W = 80,000:
W = 80,000 / L
Substituting this value of W in the perimeter equation:
P = 2L + (80,000 / L)
To minimize P, we take the derivative of P with respect to L and set it equal to zero:
dP/dL = 2 - (80,000 / L^2) = 0
Solving for L:
2L = 80,000 / L^2
2L^3 = 80,000
L^3 = 40,000
Taking the cube root of both sides:
L ≈ 34.64 meters
Now, substitute this value of L back into the equation W = 80,000 / L:
W ≈ 80,000 / 34.64 ≈ 2,310.82 meters
Therefore, to minimize the amount of fence needed, the farmer should use the dimensions of approximately 34.64 meters by 2,310.82 meters.
To minimize the amount of fence needed, we need to find the dimensions of the rectangular pasture that will give us the maximum area. The area of the rectangular pasture can be represented as length multiplied by width.
Let's say the length of the rectangular pasture is 'L' and the width is 'W'. Since the pasture lies along a river, one side will not require fencing. This means one side of the pasture will be equal to the length of the river, which we can represent as 'L_river'.
Now, we can determine the equation for the area of the pasture as follows:
Area = Length × Width
However, we have the constraint that the area of the pasture should be 80,000 square meters. So, the equation becomes:
80,000 = Length × Width
To minimize the amount of fence needed, we need to find the dimensions that will maximize the area. To do this, we can use calculus. We need to find the critical points of the area function by taking the partial derivatives with respect to both length and width and setting them equal to zero.
Let's differentiate the area equation with respect to length:
d(Area)/d(Length) = Width
Similarly, differentiate the area equation with respect to width:
d(Area)/d(Width) = Length
Setting both of these equations equal to zero, we get:
Width = 0
Length = 0
Since width and length cannot be zero (as we cannot have a zero-sized pasture), we can ignore these solutions.
Now, we need to consider the constraints given in the problem. We know that one side of the pasture is equal to the length of the river, so we have another equation:
2 × Width + Length = Length_river
So, if we substitute Length_river for Length, we get:
2 × Width + Length = Length_river
2 × Width + Length = L_river
Rearranging the equation, we get:
Length = L_river - 2 × Width
Substituting this back into the area equation:
80,000 = Length × Width
80,000 = (L_river - 2 × Width) × Width
Now, we have a quadratic equation in terms of Width. We can solve for Width using various methods, such as factoring, completing the square, or using the quadratic formula. Once we find the value of Width, we can substitute it back into the equation to find the corresponding Length.
By finding the values of Width and Length, we can determine the dimensions the farmer should use to minimize the amount of fence needed for the rectangular pasture along the river.