# Fencing next to a river

GinoJ
Posers and Puzzles 12 Mar '07 02:49
1. 12 Mar '07 02:49
You have 120 meters of fencing to enclose a rectangular plot that borders on a river. If you do not fence the side along the river, find the length and width of the plot that will maximize the area. What is the largest area that can be enclosed? Show your work.
2. 12 Mar '07 02:57
Originally posted by GinoJ
You have 120 meters of fencing to enclose a rectangular plot that borders on a river. If you do not fence the side along the river, find the length and width of the plot that will maximize the area. What is the largest area that can be enclosed? Show your work.
let the two sides perpendicular to the river both be of length x. This allos the length of the one side parellel to the river to be 120-2x. Now, it is just a simple optimazation problem:
A=(120-2x)(X)
A=120x-2x^2
now we must find the derivative
dA/dx=120-4x
to maximize area, dA/dx must equal 0 so...
0=120-4x
-120=-4x
x=30
so the demensions are 30 by 60. This maximizes the area at 180 meters^2.
3. 12 Mar '07 02:59
Originally posted by TDR1
let the two sides perpendicular to the river both be of length x. This allos the length of the one side parellel to the river to be 120-2x. Now, it is just a simple optimazation problem:
A=(120-2x)(X)
A=120x-2x^2
now we must find the derivative
dA/dx=120-4x
to maximize area, dA/dx must equal 0 so...
0=120-4x
-120=-4x
x=30
so the demensions are 30 by 60. This maximizes the area at 180 meters^2.
That was fast dogg.
4. 12 Mar '07 03:01
im in AP AB Calc (sounds like a spelling class) this year and we just learned it so it was easy...got any more?
5. 12 Mar '07 03:04
Originally posted by TDR1
im in AP AB Calc (sounds like a spelling class) this year and we just learned it so it was easy...got any more?
I do but they're all easy like this one.ðŸ˜€
6. XanthosNZ
Cancerous Bus Crash
12 Mar '07 08:05
Imagine you have 100 metres of fencing that can only be fenced in a straight line. Making a corner costs you 0.5m of fence in wastage and can only be abrupt (a sharp corner, not curved at all). What is the largest area you can enclose using a regular polygon? Can you show that this is the largest area possible using any shape?
7. 12 Mar '07 09:23
Originally posted by TDR1
let the two sides perpendicular to the river both be of length x. This allos the length of the one side parellel to the river to be 120-2x. Now, it is just a simple optimazation problem:
A=(120-2x)(X)
A=120x-2x^2
now we must find the derivative
dA/dx=120-4x
to maximize area, dA/dx must equal 0 so...
0=120-4x
-120=-4x
x=30
so the demensions are 30 by 60. This maximizes the area at 180 meters^2.
you mean 1800 m² of course
8. PBE6
Bananarama
12 Mar '07 15:31
Originally posted by XanthosNZ
Imagine you have 100 metres of fencing that can only be fenced in a straight line. Making a corner costs you 0.5m of fence in wastage and can only be abrupt (a sharp corner, not curved at all). What is the largest area you can enclose using a regular polygon? Can you show that this is the largest area possible using any shape?
The area of a regular polygon is given by:

A = (n*(k/2)^2) / tan(theta/2)

where "n" is the number of sides, "k" is the edge length, and "theta" is the angle between two neighbouring lines drawn from the centre to the vertices of the polygon. In this case:

k = (100 - 0.5*n) / n (because of the wastage at each vertex)
tan(theta/2) = tan(2*pi/2*n) = tan(pi/n)

Taking the derivative of A with respect to n, setting A' = 0, and solving for n, we find that there is a local maximum at n = 8.747. Since a regular polygon has an integer number of sides, we check n = 8 and n = 9, and find that n = 9 gives the larger area. To check to see if this area is the maximum, note that the edge length is negative for n > 200, and that this function decreases for 9 < n < 200.
9. XanthosNZ
Cancerous Bus Crash
12 Mar '07 21:58
Originally posted by PBE6
The area of a regular polygon is given by:

A = (n*(k/2)^2) / tan(theta/2)

where "n" is the number of sides, "k" is the edge length, and "theta" is the angle between two neighbouring lines drawn from the centre to the vertices of the polygon. In this case:

k = (100 - 0.5*n) / n (because of the wastage at each vertex)
tan(theta/2) = tan(2*pi/2*n) = ...[text shortened]... he edge length is negative for n > 200, and that this function decreases for 9 < n < 200.
That's the first part but can you show that no non-regular polygon exists with a greater area?
10. PBE6
Bananarama
13 Mar '07 13:43
Originally posted by XanthosNZ
That's the first part but can you show that no non-regular polygon exists with a greater area?
Hmm....that's going to be tougher.
11. PBE6
Bananarama
13 Mar '07 19:41
OK, well the area of an irregular polygon is given by:

A = (1/2)*SUM((x(i), y(i)) X (x(i+1), y(i+1))

where X is the cross-product, and x(t) and y(t) are the coordinates of the vertices of the polygon. We also know that the perimeter of the polygon is given by:

P = SUM(SQRT((x(i+1)-x(i))^2 + (y(i+1)-y(i))^2)

and this is equal to 100-0.5*n. The function P acts as a constraint on the position of the vertices. Now, I seem to recall a method for solving functions that are constrained by other functions, something to do with calculus and lambdas. Anyone help me out here? Google searching has been somewhat fruitless. My idea anyway is to find the relationship between the general solution for the area and the constraints, and prove that the optimal positioning of the vertices is equidistant from the centre of the polygon.
12. 13 Mar '07 23:29
Is this one of those rivers that runs in a very straight line like a canal type of river?
13. XanthosNZ
Cancerous Bus Crash
14 Mar '07 00:05
No, actually it's curved like a sin wave with a period of 40 metres and an amplitude of 10m (this is the equation of the bank that the fence is to be used against).
14. 14 Mar '07 00:08
Originally posted by XanthosNZ
No, actually it's curved like a sin wave with a period of 40 metres and an amplitude of 10m (this is the equation of the bank that the fence is to be used against).
I had a feeling it wasnt going to be simple
15. 14 Mar '07 08:56
Originally posted by PBE6
Now, I seem to recall a method for solving functions that are constrained by other functions, something to do with calculus and lambdas. Anyone help me out here? Google searching has been somewhat fruitless. My idea anyway is to find the relationship between the general solution for the area and the constraints, and prove that the optimal positioning of the vertices is equidistant from the centre of the polygon.
I think you're looking for Lagrange multipliers.

To find the maximum of f(x), subject to the constraint g(x) = c,

Maximise F(x, lambda) = f(x) - lambda*[g(x) - c]

But I'm not planning on trying the calculation myself!