A rectangular field is to be enclosed by a fence and divided into two smaller plots by a fence parallel to one of the side. Find the dimensions of the largest such field if 1200 m of fencing material is available. What is the area of this field and what are the dimensions that will give the largest area?
Well, I must say, this sounds like a fence-tastic problem! Let's break it down.
So, we have a rectangular field that needs to be enclosed by a fence. We also need to divide the field into two smaller plots with a fence parallel to one of the sides.
Let's call the length of the field L and the width W.
To enclose the field, we need to use fencing material for all four sides. That gives us a total of 2L + 2W.
To divide the field, we need to use fencing material for the length of the field, which is L, and the width of the field, which is W.
So, the total amount of fencing material we need is 2L + 2W + L + W. And we know that this amount is equal to 1200m.
Now, let's simplify the equation. 2L + L + 2W + W = 1200.
Combining like terms, we get 3L + 3W = 1200.
Dividing both sides by 3, we get L + W = 400.
Now, we need to find the dimensions that will give the largest area. The area of a rectangle is given by A = L * W.
So, we want to maximize the area A = L * W, while satisfying the equation L + W = 400.
To find the dimensions that give the largest area, we can use a little trick called calculus.
Using calculus, we can find that the dimensions that give the largest area are L = 200 and W = 200. In other words, the field should be a square!
So, the largest field would have dimensions of 200m by 200m, and the area would be 200m * 200m = 40,000m^2.
I hope my juggling of numbers has amused you!
To find the dimensions of the largest field, we need to maximize the area. Let's denote the length of the field as L and the width as W.
Step 1: Set up the equation for perimeter:
Since the field is to be enclosed by a fence and divided into two smaller plots by a fence parallel to one of the sides, we have:
Perimeter = 2(L + W) + L.
Given that 1200 m of fencing material is available, we can set up the equation:
2(L + W) + L = 1200.
Step 2: Simplify the equation:
Expanding the equation, we get:
2L + 2W + L = 1200,
3L + 2W = 1200.
Step 3: Solve for W in terms of L:
Subtracting 3L from both sides, we get:
2W = 1200 - 3L.
Dividing both sides by 2, we simplify:
W = (1200 - 3L) / 2.
Step 4: Formulate the equation for the area:
The area of the field is given by the product of length and width:
A = L * W.
Substituting the value of W from step 3, we have:
A = L * [(1200 - 3L) / 2].
Simplifying, we get:
A = (1200L - 3L^2) / 2.
Step 5: Find the maximum area:
To find the maximum area, we can differentiate the area equation with respect to L and set it equal to zero:
dA/dL = (1200 - 6L) / 2 = 0.
Simplifying, we find:
1200 - 6L = 0.
Solving for L, we get:
6L = 1200,
L = 200 m.
Step 6: Find the dimensions that will give the largest area:
Substituting the value of L into the equation for W from step 3, we have:
W = (1200 - 3 * 200) / 2,
W = (1200 - 600) / 2,
W = 300 / 2,
W = 150 m.
Step 7: Calculate the area with the dimensions from step 6:
Substituting the values of L and W, we have:
A = L * W,
A = 200 * 150,
A = 30000 square meters.
So, the area of the field is 30000 square meters, and the dimensions that will give the largest area are a length of 200 meters and a width of 150 meters.
To find the dimensions of the largest field, we need to determine the dimensions that will result in the largest possible area.
Let's assume that the length of the rectangular field is L and the width is W.
We are given that the total amount of fencing material available is 1200 meters. The total length of fencing required will be equal to the perimeter of the rectangular field.
The perimeter of the rectangular field is given by:
Perimeter = 2L + 3W
Since the fence parallel to one side divides the field into two smaller plots, we need to consider the length of only three sides, which is:
Perimeter = L + 2W
Given that the total fencing material available is 1200 meters, we can write the equation:
L + 2W = 1200
Now, we need to express L in terms of W, so that we can substitute it back into the equation for area.
Rearranging the above equation, we get:
L = 1200 - 2W
The area of the rectangular field is given by:
Area = L * W
Substituting the value of L, we get:
Area = (1200 - 2W) * W
To find the dimensions that will give the largest area, we need to maximize the function Area = (1200 - 2W) * W.
To find the maximum value, we can take the derivative of the area function with respect to W and equate it to zero:
d(Area)/dW = 1200 - 4W = 0
Solving this equation, we find that W = 300.
Substituting this value back into the equation for L, we get:
L = 1200 - 2W = 1200 - 2(300) = 1200 - 600 = 600
Therefore, the dimensions of the largest field that can be enclosed by 1200 meters of fencing material is 600m x 300m.
To find the area of this field, substitute the values of L and W into the equation for area:
Area = L * W = 600 * 300 = 180,000 square meters.
Hence, the area of the field is 180,000 square meters and the dimensions of the largest such field are 600m x 300m.
Draw a diagram. There will be 3 strips of one length, and 2 strips of the other. If we call them x and y, then
3x+2y=1200
The area is
A = xy=x(1200-3x)/2 = 600x-3/2 x^2
The maximum area occurs at the vertex of the parabola, at
x = -b/2a = 200
So, there will be 3 lengths of 200
and 2 lengths of 300
The maximum area is 60,000 m^2
As with all such problems, the maximum area is achieved when the fencing is divided equally between lengths and widths, no matter how many of each there are.
In this case, 1200/2 = 600
since there are 3 lengths (x), each is 200
There are 2 widths (y), each is 300