[Kravez]

Maximise P = 3x + y

Alright so how do you figure out the coordinates for the objective line from P = 3x + y

[ztibor]

(Original post by Kravez)
In the textbook it doesn't say how you will work out values for your objective or profit line.

For example:




How will you work out the profit line for P = 2x + y?

[Kravez]

Still looking.

[TenOfThem]

The objective line is 2x+y = c

[Kravez]

(Original post by TenOfThem)
The objective line is 2x+y = c

I'm not looking for the answer, I'm asking how you would you figure out the coordinates for the objective line?

Eg) http://www.edexcel.com/migrationdocu...e_20110520.pdf

Question 3.

How would you get the coordinates for the objective line? Re-arranging won't give you them

[TenOfThem]

I don't understand what you are asking

And I do not appreciate the tone, you asked for my help

[Kravez]

(Original post by TenOfThem)
I don't understand what you are asking

And I do not appreciate the tone, you asked for my help

My apologies, that wasn't aimed at you. I've edited my first post to make it more clear

[BabyMaths]

You can draw any objective line (with the correct gradient). It's the gradient that matters.

[Kravez]

(Original post by BabyMaths)
You can draw any objective line (with the correct gradient). It's the gradient that matters.

So for the question in my first post

P = 3x + y

you re-arrange to get

y = -3x + p

then you plot a line of y = -3x?

[TenOfThem]

(Original post by BabyMaths)
You can draw any objective line (with the correct gradient). It's the gradient that matters.

This is the case

I get students to set up one

for 2x+y you could start at 2x+y = 8 for example

We put our ruler as that line and then slide the ruler ... keeping the gradient ... until you hit the appropriate vertex

[JJMills]

(Original post by Kravez)
Maximise P = 3x + y

Alright so how do you figure out the coordinates for the objective line from P = 3x + y

If you re-arrange you get y = 3x - P, how are you supposed to plot this onto the graph when you can't figure out the coordinates?

You can take any value of P to plot a line. I've always been taught to take a value which is both a multiple of the coefficient of x and the coefficient of y.

In this case you could take 3 and make that P then plot the points on the graph for when x = 0 and then for when y = 0. Connect these and you'll have your objective line.

[Kravez]

(Original post by JJMills)
You can take any value of P to plot a line. I've always been taught to take a value which is both a multiple of the coefficient of x and y.

In this case you could take 3 and make that P then plot the points on the graph for when x = 0 and then for when y = 0. Connect these and you'll have your objective line.

So you would replace P by 3 giving:

3 = 3x + y

re-arrange

y = -3x + 3

x = 0: y = 3 (0, 3)

y = 0, x = 1 (1, 0)

plot the points then join the line up?

[Kravez]

(Original post by TenOfThem)
This is the case

I get students to set up one

for 2x+y you could start at 2x+y = 8 for example

We put our ruler as that line and then slide the ruler ... keeping the gradient ... until you hit the appropriate vertex

I understand how to achieve the values at the vertexs. Although why did you pick the number 8? Where did it come from? What I am trying to learn is how you achieve the profit line from lets say e.g.

P = 3x + y

P = 60x + 20y

How would you figure out the points for these two?

[JJMills]

(Original post by Kravez)
So you would replace P by 3 giving:

3 = 3x + y

re-arrange

y = -3x + 3

x = 0: y = 3 (0, 3)

y = 0, x = 1 (1, 0)

plot the points then join the line up?

Correct.

The reason you can put any value of P in is because with the rearranged formula of y = -3x + P you can see that P doesn't affect the gradient. This means that for any value of P the line you plot will be parallel with the line of any other value of P.

(Original post by Kravez)
What I am trying to learn is how you achieve the profit line from lets say e.g.

P = 3x + y

P = 60x + 20y

How would you figure out the points for these two?

As I said above, take a multiple of the two coefficients for your value of P.

For the first equation you could use 3, 6, 9, 12, etc.
For the second equation you could use 60, 120, 180, 240, etc.

This is just to make it easy to plot. It really doesn't matter what value of P you take but the above ones I mentioned make it easier.

[TenOfThem]

(Original post by Kravez)
I understand how to achieve the values at the vertexs. Although why did you pick the number 8? Where did it come from? What I am trying to learn is how you achieve the profit line from lets say e.g.

P = 3x + y

P = 60x + 20y

How would you figure out the points for these two?

As said ... you just start with an objective line

There are infinite OLs they give the Objective with different (x,y)

For P=60x+20y I would just start with 60x+20y = 120

[Kravez]

(Original post by JJMills)
Correct.

The reason you can put any value of P in is because with the rearranged formula of y = -3x + P you can see that P doesn't affect the gradient. This means that for any value of P the line you plot will be parallel with the line of any other value of P.

Alright for this question

Maximise P = 60x + 20y



They have used the points (0, 60) and (20,0)

If I took P to be 60 and got the following:

60 = 60x + 20y

3 = 3x + y
y = -3x + 3

x = 0: y = 3 (0,3)

y = 0: x = 1 (1,0)

Will this be correct? (Even though the line would be very small)

[Kravez]

...

[JJMills]

(Original post by Kravez)
Alright for this question

Maximise P = 60x + 20y



They have used the points (0, 60) and (20,0)

If I took P to be 60 and got the following:

60 = 60x + 20y

3 = 3x + y
y = -3x + 3

x = 0: y = 3 (0,3)

y = 0: x = 1 (1,0)

Will this be correct?

Yes it would be correct but you also have to factor in the scale of the graph. You don't really want to be plotting a x value of 1 on a graph that goes up in increments of 10.

Try going for P = 600 or 1200 so your points are (10, 0) and (0, 30) or (20, 0) and (0, 60).

[TenOfThem]

(Original post by Kravez)
Alright for this question

Maximise P = 60x + 20y



They have used the points (0, 60) and (20,0)

If I took P to be 60 and got the following:

60 = 60x + 20y

3 = 3x + y
y = -3x + 3

x = 0: y = 3 (0,3)

y = 0: x = 1 (1,0)

Will this be correct? (Even though the line would be very small)

The trouble is you are trying to establish the gradient so that you can move the objective line to the point ... yours would be useless


You do not have to draw a line you know .... just use the gradient ... I think that you think there is aline that forms part of the answer ... there isn't

[Kravez]

(Original post by TenOfThem)
The trouble is you are trying to establish the gradient so that you can move the objective line to the point ... yours would be useless


You do not have to draw a line you know .... just use the gradient ... I think that you think there is aline that forms part of the answer ... there isn't

Yes I realise that you don't need a profit line to work out the values and that you can use simultaneous equations to get the coordinates of the vertexs.

Although the thing is, some questions specifically ask for a objective line rather than using point testing to work out the answer that is why I want to know how to work out the profit line so if the question does pop up in the exam to specifically use a profit line then I will just write it down and use simultaneous equations to work out the values

It's all done now as I know how to do it