A farmer with 10000 meters of fencing wants to enclose a rectangular field and divide it into two plots with a fence parallel to the sides. What is the largest area that can be enclosed?
He has 10 km of fence to fence in a field on a farm ??
Anyway ...
Let each of the lengths be y
let each of the shorter sides by x
so we need 2y + 3x
2y + 3x = 10000 ---> y = (10000-3x)/2
area = xy
= x(10000-3x)/2
= 5000x - (3/2)x^2
d(area)/dx = 5000-3x
= 0 for a max of area
3x = 5000
x = 5000/3 m or 1666 2/3 m
sub that into area
then area = 4,166,666.667 m^2
lika sum bodee
Will you do this same problem but the farmer has 400m of fencing.
Ah, the wandering farmer and his fencing dilemma! To find the largest area that can be enclosed, we need to apply some mathematical clownery. Let's call one side of the rectangular field 'x' and the other side 'y'.
Now, we know that the farmer has 10,000 meters of fencing. If we add up all four sides of the rectangular field, we get:
x + y + x + y = 10,000
This simplifies to:
2x + 2y = 10,000
Dividing both sides by 2 gives us:
x + y = 5,000
Now, we need to express one side in terms of the other so we can maximize the area. Let's solve for 'y':
y = 5,000 - x
We can then use this expression for 'y' and the formula for area to find the largest enclosed area. The area is given by:
A = x * y
Replacing 'y' with '5,000 - x', we have:
A = x * (5,000 - x)
To maximize the area, we need to find the value of 'x' that makes A as big as possible. Sounds like a circus act, doesn't it? But fear not, for we can use calculus to find the maximum!
Let's take the derivative of A with respect to 'x' and set it equal to zero:
dA/dx = 5,000 - 2x = 0
Solving for 'x', we find:
2x = 5,000
x = 2,500
Substituting this value back into our expression for 'y', we get:
y = 5,000 - 2,500 = 2,500
Thus, the largest area that can be enclosed is:
A = x * y = 2,500 * 2,500 = 6,250,000 square meters
So, the wandering farmer can maximize his clownish farm area to 6,250,000 square meters. Now that's a farm that even the circus would envy!
To find the largest area that can be enclosed, we need to optimize the dimensions of the rectangle. Let's break down the problem into steps:
Step 1: Define the variables
Let's say the width of the rectangle is x meters. Since there are two equal plots, we can consider each plot as a separate rectangle, each with a width of x meters.
Step 2: Determine the equation for the perimeter
The perimeter of the rectangle is given by the sum of all sides, which includes the length, width, length, and width again. Mathematically, it can be represented as:
Perimeter = Length + Width + Length + Width = 2 * (Length + Width)
Given that the farmer has 10000 meters of fencing, we can write the equation as:
2 * (Length + Width) = 10000
Step 3: Isolate a variable
To make calculations easier, let's isolate the length variable in terms of the width:
Length = (10000 - 2 * Width) / 2
Step 4: Determine the equation for the area
The area of the rectangle is given by the product of the length and width:
Area = Length * Width
Substituting the expression for length obtained in step 3:
Area = [(10000 - 2 * Width) / 2] * Width
Step 5: Maximize the area
To find the maximum area, we take the derivative of the area equation with respect to the width and set it equal to zero. This will help us find the critical points, where the maximum occurs.
d(Area) / d(Width) = 0
Differentiating the area equation:
d(Area) / d(Width) = [(10000 - 2 * Width) / 2] - Width = 0
Simplifying the equation:
(10000 - 2 * Width) - 2 * Width = 0
10000 - 4 * Width = 0
4 * Width = 10000
Width = 10000 / 4
Width = 2500 meters
Step 6: Calculate the length and area
Using the width value obtained in step 5, we can find the length and area:
Length = (10000 - 2 * Width) / 2 = (10000 - 5000) / 2 = 2500 meters
Area = Length * Width = 2500 * 2500 = 6,250,000 square meters
Therefore, the largest area that can be enclosed with 10,000 meters of fencing is 6,250,000 square meters.