This discussion is closed.
Albuk
Badges: 0
#1
Report Thread starter 17 years ago
#1
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!
0
theone
Badges: 0
Rep:
?
#2
Report 17 years ago
#2
I guess this is a problem where you have to maximise the area of a field given a certain area of fencing correct?
0
Albuk
Badges: 0
#3
Report Thread starter 17 years ago
#3
" 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."

:confused:
0
Toffee
Badges: 1
Rep:
?
#4
Report 17 years ago
#4
Since when are there women farmers?
0
Albuk
Badges: 0
#5
Report Thread starter 17 years ago
#5
I guess it's just being politcally correct. lol :rolleyes:
0
king of swords
Badges: 0
Rep:
?
#6
Report 17 years ago
#6
(Original post by 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."

:confused:
A circle should create the largest area?
0
theone
Badges: 0
Rep:
?
#7
Report 17 years ago
#7
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.
0
PQ
Badges: 21
Rep:
?
#8
Report 17 years ago
#8
(Original post by 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."

:confused:
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)
0
theone
Badges: 0
Rep:
?
#9
Report 17 years ago
#9
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.
0
PQ
Badges: 21
Rep:
?
#10
Report 17 years ago
#10
(Original post by 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.
0
Toffee
Badges: 1
Rep:
?
#11
Report 17 years ago
#11
(Original post by 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.
Isn't this GCSE maths we're talking about here?
0
king of swords
Badges: 0
Rep:
?
#12
Report 17 years ago
#12
(Original post by 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.
0
theone
Badges: 0
Rep:
?
#13
Report 17 years ago
#13
(Original post by 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.
0
Ralfskini
Badges: 10
Rep:
?
#14
Report 17 years ago
#14
Surely its a choice between calculus or trial and error?!
0
Albuk
Badges: 0
#15
Report Thread starter 17 years ago
#15
(Original post by theone)
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!
0
theone
Badges: 0
Rep:
?
#16
Report 17 years ago
#16
(Original post by Albuk)
first principles...?

Btw thanks for all your help so far everyone!
I mean you need not know any knowledge of calculus to deduce what I've written, you can deduce it from what you yourself already should know at GCSE higher tier.
0
Ralfskini
Badges: 10
Rep:
?
#17
Report 17 years ago
#17
(Original post by 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)
What's the point of doing an equlateral triangle, circle, etc because the area is fixed and determined by the perimeter/ circumference?
0
theone
Badges: 0
Rep:
?
#18
Report 17 years ago
#18
(Original post by Ralfskini)
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.
0
Ralfskini
Badges: 10
Rep:
?
#19
Report 17 years ago
#19
(Original post by 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.
0
theone
Badges: 0
Rep:
?
#20
Report 17 years ago
#20
(Original post by Ralfskini)
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.
Well this is only true when the perimeter fixes all other parameters.
0
X
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Following the government's announcement, do you think you will be awarded a fair grade this year?

Yes (527)
51.46%
No (497)
48.54%

Watched Threads

View All
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise