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edexcel d1 objective line

I dont understand how to draw the objective line?I know that u make the function equal to a common multiple,but in different papers they draw the lines differently and...i just dont get it.How should I know which points to choose?
Here is an example,question 7 https://qualifications.pearson.com/content/dam/pdf/A%20Level/Mathematics/2013/Exam%20materials/6689_01_rms_20120816.pdf
Original post by runballadmix
I dont understand how to draw the objective line?I know that u make the function equal to a common multiple,but in different papers they draw the lines differently and...i just dont get it.How should I know which points to choose?
Here is an example,question 7 https://qualifications.pearson.com/content/dam/pdf/A%20Level/Mathematics/2013/Exam%20materials/6689_01_rms_20120816.pdf


What is the objective function? It's best to give the method on the relevant function here.
Original post by RDKGames
What is the objective function? It's best to give the method on the relevant function here.

80x+80y=P
Original post by runballadmix
80x+80y=P


Ok, so for a function like that, first of all choose any sensible value of PP.

In the mark scheme, they chose P=800P=800 (even though they don't say it) which means we have 80x+80y=80080x+80y = 800. This simplifies to x+y=10x+y = 10.

Now mark down the x and y intercepts of this function. When x=0x=0 we have y=10y = 10. Hence mark down (0,10).
When y=0y=0 we have x=10x=10. Hence mark down (10,0)(10,0).

Now join these two points up with a dashed line and label it as the objective function.


As I said, you can choose any PP you like but with practice you can start to notice which values are sensible choices. It doesn't matter which P you choose because at the end of the day, you're going to be sliding that line across on the graph which is equivalent to the value of P constantly changing in the first place; all you're doing is choosing a specific value and sketching the objective function to see how it looks like. So you're always going get the same optimal solution.
Original post by RDKGames
Ok, so for a function like that, first of all choose any sensible value of PP.

In the mark scheme, they chose P=800P=800 (even though they don't say it) which means we have 80x+80y=80080x+80y = 800. This simplifies to x+y=10x+y = 10.

Now mark down the x and y intercepts of this function. When x=0x=0 we have y=10y = 10. Hence mark down (0,10).
When y=0y=0 we have x=10x=10. Hence mark down (10,0)(10,0).

Now join these two points up with a dashed line and label it as the objective function.


As I said, you can choose any PP you like but with practice you can start to notice which values are sensible choices. It doesn't matter which P you choose because at the end of the day, you're going to be sliding that line across on the graph which is equivalent to the value of P constantly changing in the first place; all you're doing is choosing a specific value and sketching the objective function to see how it looks like. So you're always going get the same optimal solution.

thank you so much!So in the exam,it doesn't matter what P or what values of x and y i choose,right?
Original post by runballadmix
thank you so much!So in the exam,it doesn't matter what P or what values of x and y i choose,right?


It doesn't matter what P you choose (as long as it's not zero), and technically it doesn't matter what x,yx,y you choose to go along with it but I hope you can see how the x and y intercepts are obviously simpler to determine than any other point on the line.

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