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Hi all,

I am really stuck with how to do this! I have had a search around, but basically what I need is like a plan of what to do. Ie. ....Where do I start?!

I am doing GCSE Higher tier, but have no idea! I missed the last week of school, so that just adds to the confusion!

Any help, ideas, plans, hints ot tips would be REALLY appreciated!

Thanks in advance!

I am really stuck with how to do this! I have had a search around, but basically what I need is like a plan of what to do. Ie. ....Where do I start?!

I am doing GCSE Higher tier, but have no idea! I missed the last week of school, so that just adds to the confusion!

Any help, ideas, plans, hints ot tips would be REALLY appreciated!

Thanks in advance!

Scroll to see replies

" 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."

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."

Albuk

" 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."

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 circle should create the largest area?

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.

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.

Albuk

" 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."

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."

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)

theone

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.

I'd say a combination of the 2 (ie working out the relationship between l & a for each different number of sides) would be a dead cert to get extremely high marks.

theone

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.

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?

Toffee

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.

king of swords

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.

Pencil Queen

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)

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?

theone

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.

- NEED Help deciding my a levels
- GCSE English Literature Study Club: An Inspector Calls
- English GCSE revision help 2024
- City of London application unsuccessful
- GCSE English literature paper 1 help
- suhan's gyg for 2024 !
- English Gcse's
- My gsce *cramming* study sessions🔝🌊
- English
- Bouts of depression . Affecting me academically and my social life
- chances for oxbridge base on gcse
- Could anyone say what I need to improve on/what marks I'd get for English P1 Q2?
- Any high-achievers willing to share revision techniques for GCSEs?
- Should I take my GCSE’s this year or next year?
- Help me is this ok in an exam
- how to improve GSCE ENGLISH grades
- last minute yr 10 mock revision HELP!!
- revison
- I am worried of my grades will affect my chances in future for a law firm
- Paris 2024: Olympics and Paralympics - ALL DISCUSSIONS AND GOOD STUFF wooooooo

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