- 12 Mar '07 02:57

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:*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.**

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. - 12 Mar '07 02:59

That was fast dogg.*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. - 12 Mar '07 08:05Imagine 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?
- 12 Mar '07 09:23

you mean 1800 m² of course*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. - 12 Mar '07 15:31

The area of a regular polygon is given by:*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?**

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. - 12 Mar '07 21:58

That's the first part but can you show that no non-regular polygon exists with a greater area?*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. - 13 Mar '07 19:41OK, 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. - 14 Mar '07 08:56

I think you're looking for Lagrange multipliers.*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.**

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!