A farmer has 120 feet of fencing to enclose a rectangular plot for some of his animals. One side of the area borders on a barn.
a.) If the farmer does not fence the side along the barn, find the length and width of the plot that will maximize the area?
b.) What is the maximum area that can be enclosed?
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I'm not sure if this is even close to how part a) should start but is it 2x+y=120? I'm confused.
let the length parallel to the barn be y ft
let the two other sides be x ft each
so 2x + y = 120
y = 120 - 2x
area = xy
= x(120-2x)
= -2x^2 + 120x
this is a downward parabola,
the x of the vertex is -b/2a = -120/-4 = 30
then y = 120-60 = 60
So the largest area is 30(60) or 1800 ft^2
when the long side is 60 ft and the short side is 30 ft
check:
available fence = 120 ft
one long side + 2 short sides = 60 + 2(30) = 120
suppose we use 29 ft for the short side, then the longer side is 120-2(29) = 62
and the area is 29(62) = 1798 which is less than 1800
suppose we use 31 ft for the short side, then the longer side is 120-2(31) = 58
and the area is 31(58) = 1798 which is less than 1800
To solve this problem, we can use the concept of optimization. Let's break down the problem step by step:
Step 1: Define the variables
Let's assume the length of the plot is represented by "x" feet, and the width of the plot (perpendicular to the barn) is represented by "y" feet.
Step 2: Determine the constraints
We are given that the farmer has 120 feet of fencing. Since one side borders the barn and doesn't need to be fenced, the total amount of fencing required is the same as the perimeter of the rectangular plot, excluding the side next to the barn:
Perimeter = 2x + y = 120
Step 3: Express the area of the plot
The area of a rectangle is given by the formula:
Area = length × width
Area = x × y
Step 4: Solve for y in terms of x
From the constraint equation in Step 2, we can express y in terms of x:
y = 120 - 2x
Step 5: Substitute y in terms of x into the area equation
Now, substitute the expression for y into the area equation to obtain the area in terms of x:
Area = x × (120 - 2x)
Area = 120x - 2x^2
Step 6: Maximize the area
To find the maximum area, we need to find the critical points of the area equation. Firstly, take the derivative of the area equation with respect to x:
d(Area)/dx = 120 - 4x
Set the derivative equal to zero and solve for x:
120 - 4x = 0
4x = 120
x = 30
Step 7: Find the corresponding value of y
Substitute the value of x back into the equation for y to find its value:
y = 120 - 2x = 120 - 2(30) = 120 - 60 = 60
Step 8: Calculate the maximum area
Substitute the values of x and y into the area equation to find the maximum area:
Area = x × y = 30 × 60 = 1800 square feet
Therefore, the length and width of the plot that will maximize the area are 30 feet and 60 feet, respectively, and the maximum area that can be enclosed is 1800 square feet.
To solve the problem, let's start by defining variables for the length and width of the rectangular plot. Let's call the length x and the width y.
a) If the farmer does not fence the side along the barn, the fence will have three sides with a total length of 120 feet. The formula to calculate the perimeter of a rectangle is P = 2x + 2y.
Since we are not fencing the side along the barn, the perimeter is given by the equation 2x + y = 120. The barn essentially acts as one side of the rectangle.
To maximize the area of the plot, we need to express the area of the rectangle in terms of a single variable. The area of a rectangle is given by A = xy.
Now, we can use the equation 2x + y = 120 to find y in terms of x. Rearrange the equation to solve for y:
y = 120 - 2x
Substitute this value of y into the area equation:
A = x(120 - 2x)
Simplify the equation:
A = 120x - 2x^2
To find the maximum area, we need to find the value of x that maximizes the equation A = 120x - 2x^2. We can do this by taking the derivative of A with respect to x, setting it equal to zero, and solving for x.
dA/dx = 120 - 4x = 0
Solve for x:
120 - 4x = 0
4x = 120
x = 30
Now that we have the value of x, we can substitute it back into the equation y = 120 - 2x to find y:
y = 120 - 2(30)
y = 120 - 60
y = 60
So, the length of the plot that will maximize the area is 30 feet, and the width is 60 feet.
b) To find the maximum area, substitute the values of x and y into the area equation:
A = xy
A = 30 * 60
A = 1800 square feet
Therefore, the maximum area that can be enclosed is 1800 square feet.