A farmer is constructing a rectangular pen with one addtional fence across its width.find the maximum area that can be enclosed with 2400m of fencing .
3x+2y = 2400
a = xy = x(2400-3x)/2
= 1200x - 3/2 x^2
this is a parabola, with vertex at (400,240000)
as usual, note that the fencing is evenly divided between widths and lengths.
To find the maximum area that can be enclosed with the given amount of fencing, we need to determine the dimensions of the rectangular pen.
Let's assume the length of the pen is L and the width is W. Since there are two lengths and three widths (including the additional fence), the total amount of fencing required can be calculated as follows:
2L + 3W = 2400
Now, let's solve this equation for L:
2L = 2400 - 3W
L = (2400 - 3W) / 2
Substituting this value of L back into the equation, we get:
2((2400 - 3W) / 2) + 3W = 2400
2400 - 3W + 3W = 2400
2400 = 2400 (This implies that any value of W will work)
Now, we can calculate the area of the rectangular pen using the formula:
Area = Length * Width
Area = ((2400 - 3W) / 2) * W
Area = (2400W - 3W^2) / 2
To find the maximum area, we can differentiate the area formula with respect to W:
d(Area) / dW = (2400 - 6W) / 2
Setting this derivative equal to zero gives us:
2400 - 6W = 0
6W = 2400
W = 400
Now, substitute this value of W back into the equation for L:
L = (2400 - 3(400)) / 2
L = (2400 - 1200) / 2
L = 1200 / 2
L = 600
Therefore, the dimensions of the rectangular pen that maximize the area are 600m in length and 400m in width. The maximum area that can be enclosed with 2400m of fencing is 240,000 square meters.
To find the maximum area that can be enclosed with 2400m of fencing, we need to determine the dimensions of the rectangular pen.
Let's denote the length of the rectangular pen as L and the width as W.
We know that the farmer is using 2400m of fencing, which means the total length of the four sides of the pen will be 2400m.
The perimeter (P) of the rectangular pen can be expressed as:
P = 2L + W
Since the farmer is adding an additional fence across the width, the width would be divided into two equal parts. Therefore, the width can be expressed as:
W = W1 + W2
Where W1 and W2 are the two equal parts of the width.
The perimeter equation can be rewritten as:
P = 2L + W1 + W2
Given that P = 2400, we have:
2400 = 2L + W1 + W2
To maximize the area, we need to find the dimensions that maximize the length and the width.
Since the perimeter equation has three variables (L, W1, W2), we need another equation to solve this system. The additional equation can be derived from the area (A) of the rectangle, which is given by:
A = L * W
Now, let's solve for one variable in terms of the other variables so we can substitute it into the area equation.
From the perimeter equation, we can isolate L:
2L = 2400 - W1 - W2
L = (2400 - W1 - W2) / 2
Substituting L into the area equation:
A = ((2400 - W1 - W2) / 2) * W
To find the maximum area, we can take the derivative of the area equation with respect to either W1 or W2 and set it equal to zero. However, since W1 and W2 are equal (as it is divided into two equal parts), we can solve for one variable only.
Let's take the derivative of the area equation with respect to W:
dA/dW = (2400 - W - W) / 2 = (2400 - 2W) / 2 = 1200 - W
Setting dA/dW equal to zero to find the stationary point:
1200 - W = 0
W = 1200
Substituting W = 1200 back into the perimeter equation:
2400 = 2L + 1200 + 1200
2400 = 2L + 2400
2L = 2400 - 2400
2L = 0
L = 0
From the above result, we found that L = 0, which is not a meaningful solution. Therefore, an optimal rectangular pen cannot be formed with 2400m of fencing.
Hence, we cannot enclose a maximum area with 2400m of fencing using a rectangular pen.