A rectangular field is enclosed by a fence and separated into two parts by a fence parallel to one of the sides. If 600 m of fencing is available, what should the dimensions be so that the area is a maximum?
t=56
To find the dimensions that will maximize the area, we can use calculus.
Let's denote the width of the rectangle as x and the length as y.
From the given information, we know that the rectangle is divided into two parts by a fence parallel to one of the sides. Therefore, the rectangle can be represented as follows:
x
+----+----+ y
| |
+----+----+
We are given that the total length of fencing available is 600 m, which means the perimeter of the rectangle can be expressed as:
2x + y + x = 600
Combining like terms, we get:
3x + y = 600
To solve this equation for y, we can isolate y:
y = 600 - 3x
Now, we can express the area of the rectangle as a function of x:
A = x * y
Substituting the value of y, we have:
A = x * (600 - 3x)
To find the maximum area, we need to find the critical points of this function. Taking the derivative of A with respect to x and setting it equal to zero, we have:
dA/dx = 600 - 6x = 0
Solving for x, we get:
600 - 6x = 0
6x = 600
x = 100
Now that we have the value of x, we can substitute it back into the equation of the perimeter to find the value of y:
3x + y = 600
3 * 100 + y = 600
300 + y = 600
y = 600 - 300
y = 300
Therefore, the dimensions of the rectangle that maximize the area are 100 m for the width and 300 m for the length.
To find the dimensions that maximize the area of the rectangular field, we can first express the area as a function of a single variable and then find the maximum value of that function.
Let's assume the length of the rectangular field is "x" and the width is "y". Since the field is separated into two parts by a fence parallel to one of the sides, we can express the perimeter of the field as:
2x + 3y = 600
Here, we consider the two lengths of x and the three widths of y because there are two vertical sides and three horizontal sides (two on the sides and one in the middle to separate the two parts of the field).
To express the area of the field, we use the formula:
Area = length × width
Area = x × y
Now, we have two variables and two equations. We can solve these equations to find the values of x and y that maximize the area of the field.
From the first equation, we can express x in terms of y:
2x = 600 - 3y
x = (600 - 3y)/2
Substituting this value of x in the area equation:
Area = [(600 - 3y)/2] × y
Area = (600y - 3y^2)/2
To find the maximum area, we need to find the vertex of the quadratic equation (600y - 3y^2)/2. The x-coordinate of the vertex can be found using the formula:
x = -b/(2a)
In our case, a = -3/2 and b = 600/2 = 300.
x = -300 / (2 * (-3/2))
x = -300 / (-3)
x = 100
Once we have the x-coordinate of the vertex, we can substitute it back into the equation to find the y-coordinate:
Area = (600(100) - 3(100)^2)/2
Area = (60000 - 30000)/2
Area = 30000/2
Area = 15000
Therefore, the maximum area of the rectangular field is 15000 square meters. To achieve this maximum area, the dimensions of the field should be x = 100 meters and y = y.