if a farmer has 100 feet of fence and wants to make a rectangular pigpen, one side of which is along existing straight fence.What dimensions should be used in order to maximize the area of the pen?
25 - 25 - 50
Maximum Area = 1250 ft^2
How to get that answer:
Let x be the side length that is used twice. The side parallel to the existing fence then has length 100 - 2x.
The area A is then
A(x) = x*(100 - 2x) = 100 x - 2x^2
dA/dx = 0 when
100 = 4x
x = 25
To maximize the area of the pigpen, we need to determine the dimensions that will give us the largest possible area within the given constraints. In this case, the farmer has 100 feet of fence available and one side of the pigpen will be along an existing straight fence.
Let's start by identifying the sides of the rectangular pigpen. Let's assume that the side parallel to the existing fence is x feet. This means the other two sides will be perpendicular to the existing fence and have a length of y feet each.
To calculate the area of the rectangular pigpen, we use the formula: Area = length × width = x × y.
Next, we need to consider the constraints. We know that the farmer has a total of 100 feet of fence available. Since one side of the pigpen is already covered by the existing fence, we need to account for the remaining three sides.
The remaining three sides will be made up of two perpendicular sides (2y) and the side parallel to the existing fence (x). Considering all three sides, we have:
2y + x = 100.
To find the dimensions that will maximize the area, we can solve for y in terms of x using the equation above:
2y = 100 - x
y = (100 - x)/2.
Substituting this value of y into the area formula, we have:
Area = x × ((100 - x)/2)
Area = (100x - x^2)/2.
To find the maximum area, we take the derivative of the area equation with respect to x and set it equal to zero:
d(Area)/dx = 100/2 - 2x/2 = 0
50 - x = 0
x = 50.
Therefore, the side parallel to the existing fence should be 50 feet.
To find the value of y, we substitute the value of x into the equation:
y = (100 - x)/2
y = (100 - 50)/2
y = 50/2
y = 25.
So, the dimensions that will maximize the area of the pigpen are 50 feet by 25 feet.
To maximize the area of the pigpen, you should determine the dimensions that allow for the largest possible rectangle.
Step 1: Define the variables
Let's assume the length of the rectangular pigpen (along the existing fence) is "x" feet.
The width of the pigpen (perpendicular to the existing fence) will be "y" feet.
Step 2: Formulate the equation for the perimeter
The perimeter of the pigpen is the sum of all its sides. In this case, it will be:
2x + y = 100 feet
Step 3: Solve for y in terms of x
Rearrange the equation to solve for y:
y = 100 - 2x
Step 4: Formulate the equation for the area
The area of the pigpen is given by the product of its length and width:
Area = x * y
Step 5: Substitute the value of y from Step 3 into the area equation
Area = x * (100 - 2x)
Step 6: Expand and arrange the equation
Area = 100x - 2x^2
Step 7: Find the derivative of the area equation
To find the maximum area, calculate the derivative of the area equation and set it equal to zero.
d(Area)/dx = 100 - 4x
Step 8: Solve for x
Set the derivative equal to zero and solve for x:
100 - 4x = 0
4x = 100
x = 25
Step 9: Calculate the value of y
Substitute the value of x back into the equation for y from Step 3:
y = 100 - 2x
y = 100 - 2(25)
y = 100 - 50
y = 50
Step 10: Maximize the area
The dimensions that maximize the area of the pigpen are x = 25 feet and y = 50 feet.
Therefore, the farmer should build a rectangular pigpen with dimensions 25 feet x 50 feet to maximize the area.