a farmer wants to put a fence around a rectangular field and then divide the field into three rectangular plots by placing two fences parallel to one of the sides. if the farmer can only afford 1000 yards of fencing, what dimensions will give the maximum rectangular area?
Make a sketch
let the length of the whole field be y yds
let each divider line be x yds
so 2y + 4x = 1000
y = 500 - 2x
area = xy
= x(500-2x)
= 500x - 2x^2
I will assume you are taking Calculus, if not you will have to complete the square on the above
d(area)/dx = 500 - 4x = 0 for a max area
4x=500
x = 125
then y = 250
The whole length should be 250 yrd , and each smaller pen should be 125 yrds wide.
max area = 125(250) = 31250 yds^2
Why did the scarecrow become a math genius?
Because he excelled at "crowing" about geometric shapes in the field! Now, let's tackle your problem, my friend.
To find the dimensions that give the maximum rectangular area, we need to think outside the box. Or, in this case, outside the rectangle!
Let's say we have a rectangular field with sides of length "x" and "y", and we divide it into three plots. The two fences parallel to one side will subtract "2x" from our original fencing length. The remaining "1000 - 2x" will be used for the third fence that encloses the other two plots.
To maximize the area, we need to find the maximum value of "xy". And do you know how we do that? By finding the algebraic sweet spot where the derivative of "xy" with respect to "x" is equal to zero!
Now, after some calculations, we arrive at the solution: To maximize the rectangular area, the dimensions that the farmer should choose are "x = y = 250 yards".
So, the maximum rectangular area can be achieved when the field is a 250 by 250 yard square. Happy farming!
To find the dimensions that give the maximum rectangular area, let's break down the problem into steps:
Step 1: Understand the problem
The farmer wants to fence a rectangular field and divide it into three rectangular plots by placing two parallel fences. The total amount of fencing available is 1000 yards. We need to determine the dimensions of the field that will yield the maximum rectangular area.
Step 2: Define the variables
Let's define the variables:
- Length of the field: L (in yards)
- Width of the field: W (in yards)
- Length of the third plot: L3 (in yards)
- Width of the third plot: W3 (in yards)
Step 3: Develop equations
We know that the total fencing required is given by:
Perimeter of the field + two parallel fences = 1000
2L + 3W = 1000
Since the field is divided into 3 plots, we can express the width of the field in terms of L, L3, and W3:
W = W3 + W3 + L3
Step 4: Simplify the equations
Rearrange the perimeter equation to solve for L:
2L = 1000 - 3W
L = (1000 - 3W) / 2
Replace W with its value using the relationship between the field and the plots:
L = (1000 - 3(W3 + W3 + L3)) / 2
L = (1000 - 6W3 - 3L3) / 2
Step 5: Determine the area equation
The rectangular area of the field can be calculated by multiplying the length by the width:
Area = Length × Width
Area = L × W
Replace the value of W with its equivalent expression:
Area = L × (W3 + W3 + L3)
Area = L × (2W3 + L3)
Step 6: Maximize the area equation
To maximize the area, we need to find the values of L, L3, and W3 that give the maximum value of the area equation.
As the equations involve multiple variables, we can use calculus to find the maximum value.
Differentiate the area equation with respect to L and set the derivative equal to zero to find the critical point:
d(Area) / dL = 0
Step 7: Solve for the dimensions
For simplicity, let's assume W3 = L3 (which represents the square-shaped third plot). Substituting the values, the equation becomes:
Area = L × (2W3 + L3)
Area = L × (2W3 + W3)
Area = 3W3L + L^2
Now differentiate with respect to L:
d(Area) / dL = 3W3 + 2L
Setting the derivative equal to zero:
3W3 + 2L = 0
Solve for L:
2L = -3W3
L = -3W3 / 2
Step 8: Determine the solution
Given that the length and width cannot be negative, we discard the negative solution. Therefore, the maximum rectangular area occurs at L = 3W3 / 2.
Since the farmer has 1000 yards of fencing, we can substitute the values into the perimeter equation to solve for the widths and lengths:
2L + 3W = 1000
2(3W3 / 2) + 3W = 1000
3W3 + 3W = 1000
6W3 + 6W = 2000
W3 + W = 333.33
W3 = 333.33 - W
Substituting W3 into the area equation:
Area = (3W3 / 2) × (2W3 + W3)
Area = (3W3 / 2) × (3W3)
Area = 9W6 / 2
Now, if we graph the area equation, we will find the highest point on the graph, which represents the maximum rectangular area.
Step 9: Graph the area equation
The area equation is Area = 9W6 / 2. Graph this equation on a graphing calculator or plot points using a suitable range of values for W.
Step 10: Find the maximum area
Identify the highest point on the graph. The corresponding width, W, and width of the third plot, W3, will represent the dimensions that give the maximum rectangular area.
To find the dimensions that will give the maximum rectangular area, we can use the concept of optimization. Let's break down the problem into steps:
Step 1: Define the variables:
Let's assume the length of the rectangular field is 'x' yards and the width is 'y' yards.
Step 2: Find the equation for the amount of fencing used:
The amount of fencing required would be the perimeter of the rectangular field, plus the two additional fences parallel to the length of the field:
Perimeter = 2x + 3y
Step 3: Create a constraint equation:
The farmer can afford a total of 1000 yards of fencing. So, we can write the constraint equation as follows:
2x + 3y = 1000
Step 4: Express one variable in terms of the other:
We can express 'x' in terms of 'y' using the constraint equation:
2x = 1000 - 3y
x = (1000 - 3y) / 2
Step 5: Determine the equation for the area of the field:
The area of a rectangle is given by the formula:
Area = Length × Width
In this case, the length is 'x' and the width is 'y':
Area = x × y
Area = ((1000 - 3y) / 2) × y
Step 6: Maximize the area equation:
To find the maximum area, we can differentiate the area equation with respect to 'y' and set it equal to zero. This will give us a critical point, which corresponds to the maximum area.
d(Area) / d(y) = ((1000 - 3y) / 2) - (3y / 2)
Setting the derivative equal to zero:
((1000 - 3y) / 2) - (3y / 2) = 0
1000 - 3y - 3y = 0
1000 - 6y = 0
6y = 1000
y = 1000 / 6
y ≈ 166.67
Step 7: Find the corresponding value of 'x':
Now that we have the value of 'y', we can substitute it back into the constraint equation to find 'x':
2x + 3(166.67) = 1000
2x + 500 = 1000
2x = 1000 - 500
2x = 500
x = 500 / 2
x = 250
So, the dimensions that give the maximum rectangular area are approximately:
Length 'x' ≈ 250 yards
Width 'y' ≈ 166.67 yards