GSCE Maths "The Fencing Problem"

" A farmer has exactly 1000 metres of fencing and wants to fence off a plot of level land.

She is not concerned about the shape of the plot, but it must have a perimeter of 1000m (or circumference).

She wishes to fence off the plot of land, which contains the maximum area.

Investiagate the shape, or shapes, that could be used to fence in the maximum area using exactly 1000 metres of fencing each time."

" A farmer has exactly 1000 metres of fencing and wants to fence off a plot of level land.

She is not concerned about the shape of the plot, but it must have a perimeter of 1000m (or circumference).

She wishes to fence off the plot of land, which contains the maximum area.

Investiagate the shape, or shapes, that could be used to fence in the maximum area using exactly 1000 metres of fencing each time."

Of course I've only dealt with 4 sided objects, you could do with what Pencil Queen has said and deduce the area for other n-sided polygons.

Well firstly if we have a rectangle, and have x metres as one side, we have 500-x as the other side (can you see where this comes from?). Thus the area is x(500-x) = 500x - x^2.

This next bit can be quite tricky.

Let f(x) = x^2. Let's say we increase f(x) by a small amount, call it h, so we get f(x+h) = (x+h)^2. Also by drawing a graph we can see that the gradient (call this f '(x)) is approximately (f(x+h)-f(x))/h, thisbeing the change in y over the change in x. Now expanding, we get f '(x) = (x^2 + 2xh + h^2 - x^2)/h = 2x + h.

Now if we let h become very small, so that it is essentialy 0, we can disregard the h, as it is negligable, as such the gradient of x^2 is 2x (such a process of taking h to be infinitely small is called taking a limit).

Right, now we can see that if f (x) = 500x - x^2, f '(x) = 500 - 2x.

Now for one more point, we note that if the gradient of a line is 0, then that point where the gradient is 0 must be a turning point of some form (either a maximum or a minimum or an inflexion {inflexion as in an x^3 graph at the origin). Also note that if the change in the gradient (call this f ''(x)) is less than 0, then after this point where the gradient is 0 (stick with me here) the gradient will then decrease and so our function will begin to decrease as such we see that this point where it turns is a maximum.

Ok, f ''(x) = -2. And if f '(x) = 0, then x = 250 which is a maximum. It now turns out the shape with maximal area is in fact a square with area 250^2.

Now you can look at other shapes such as circles and possibly even ellipses if you have met them.

Let me know if you don't understand anything.

Isn't this GCSE maths we're talking about here?

If a higher tier gcse maths student understands calculus then he/she can use it, but obviously that is assuming the person who started this thread DOES know some calculus.

However, what I've written here really doesn't assume knowledge of any calculus at all, it is essentially first principles.

first principles...?

Btw thanks for all your help so far everyone!

Start with 3 sided shapes (ie an equilateral triangle and a non-equilateral triangle) and work out the maximum area (ie an equilateral traingle will have sides of 333m long and the area = (height*base)/2).
Then 4 sided shapes (square, rectangle, diamond, trapezium etc) and do the same.
Then a 5 sided shape
etc etc
And of course a perfect circle (and if you can be bothered and elipse)
The equations for finding the area of these shapes should be straight forward to find (eg here (or draw scale versions on a grid and work out the areas for irregular shapes the old fashioned way)

What's the point of doing an equlateral triangle, circle, etc because the area is fixed and determined by the perimeter/ circumference?

No it isn't, since a 400x100 rectangle does not have the same area as a 250x250 square, but has the same perimeter.

That's not what I meant- you can do it with rectangles (in which case the ratio between longest and shortest side is 2:1 to maximise area) but with a square u cant change anything, nor a circle.