A farmer is making a pig pen next to the barn. He has enough materials to build 50 feet of fencing.
a)what is the largest area he can fence in?
b)what are the dimensions of the pen with the largest area?
One side of the barn counts as an enclosure.
length = 50-2w
area = w(50-2w)
max area at w=25/2
One side of the barn counts as one side of the fence. You are left with one L (length) and 2W (width). Set L=y and W=x. xy=Area(A) and 2x+y=Total Fence(50ft). Subtract 2x over to get y=50-2x. Substitute for y into the area formula to get A=x(50-2x) = A=-2x^2+50x. Use the formula x=-b/2a where a=-2 and b=50, so x=-50/(2*-2) so x=12.5ft, and y=25ft.
a) The largest area is 12.5*25=312.5ft
b)The dimensions are 12.5 by 25 ft
To find the largest area that the farmer can fence in, we need to determine the shape that maximizes the area with the given amount of fencing.
a) Let's consider a rectangular pen.
Let the length of the pen be L and the width be W.
Since the total amount of fencing material is 50 feet, we can express the perimeter of the pen as:
Perimeter = 2L + 2W
Given that Perimeter = 50 feet, we can rewrite the equation as:
2L + 2W = 50
To express the area of the pen, we use the formula for the area of a rectangle:
Area = Length × Width
We can express the area as:
Area = L × W
Now, we need to represent one of the variables (either L or W) in terms of the other. Let's solve the equation 2L + 2W = 50 for L:
2L = 50 - 2W
L = (50 - 2W) / 2
L = 25 - W
Substituting this value of L in the area equation:
Area = (25 - W) × W
Area = 25W - W^2
Now, we have the area expressed in terms of a single variable, W. To find the maximum area, we need to find the critical points of this equation. We can do this by taking the derivative of the area equation with respect to W and setting it equal to zero:
dA/dW = 25 - 2W = 0
Solving this equation, we get:
25 - 2W = 0
2W = 25
W = 25/2 = 12.5
So, the width W = 12.5 feet.
Substituting the value of W into L = 25 - W:
L = 25 - 12.5 = 12.5
Therefore, the dimensions of the pen that maximize the area are:
Length = 12.5 feet, Width = 12.5 feet.
b) The dimensions of the pen with the largest area are 12.5 feet by 12.5 feet.