a farmer can afford 8000 square meters of fence and would like to come up with the best way to use this fence, such that:
- he wants all the pens to be of the same rectangular shape.
- he wants to give each calce of his farm as much space as possible.
- he needs an easy access to each calf
he wants 16 eparate spaces.
This depends on how the pens are arranged. If the pens are all separate, then more fence is used than if they share sides.
Also, fence is not measured in square meters, but just meters.
16 pens in a line of width x and length y, sharing the walls of width x, use up
17x+32y meters of fence.
Make a double row sharing long sides y as well, and you use
18x+24y meters.
Have 16 separate pens, and you use
16(2x+2y) meters
Pin down the problem some more, and maybe we can get somewhere.
For a given perimeter, the rectangle that encloses the maximum area using one side as a building, river, fence, etc. or an adjacent section of fence, is in the ratio of 2/1. Assuming each of the 16 grazing areas is in the length to width ratio of 2/1:
Let the length of the 16 enclosed areas be L, the width of the 16 enclosed areas be W, the length of a single area is l and the length of a single area is W.
L = 16w
W/w = 2
L = 2W
17W + 2L = 8000
Using W/w = 2 and L = 16w
34w + 32w = 8000
66w = 8000
w = 121.212
W = 242.424
L = 16(121.212) = 1939.36
17(242.424) + 2(1939.36)
4121.208 + 3878.72 = 8000 meters.
To determine the optimal solution for the farmer's fence allocation, we need to calculate the dimensions of each rectangular pen that maximize the area. Here's how we can approach it:
1. Define the number of separate spaces: Given that the farmer wants 16 separate spaces, we will divide the available fence equally among these pens.
2. Determine the shape of the pens: Let's assume that the farmer wants all the pens to have a rectangular shape.
3. Calculate the maximum area for each pen: To find the dimensions that maximize the area, we can use calculus. Let's assume the length of the rectangular pen is 'l' and the width is 'w'. The perimeter of the pen is given as 8000 meters: 2l + 2w = 8000. Rearranging the equation, we have l + w = 4000.
4. Express the area in terms of one variable: We can express the area (A) of each pen as A = l * w.
5. Solve for one variable in terms of the other: Substitute l = 4000 - w into the area equation to obtain A = (4000 - w) * w.
6. Find the critical points of the area function: By taking the derivative of A with respect to w and setting it equal to zero, we can find where the area function reaches a maximum.
7. Calculate the dimensions of each pen: By solving the equation, we can find the dimensions of each pen that give the maximum area.
8. Check the number of pens: Verify if the number of pens obtained satisfies the requirement of 16 separate spaces. If not, adjust the dimensions accordingly.
To summarize, the farmer needs to divide his available fence into 16 separate rectangular pens. By calculating the dimensions that maximize the area for each pen, he can ensure that each calf gets as much space as possible.