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# D1 past paper question

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on the last question, how should I approach this?
I have found the coordinate of A and D and set up an equation so that:
the P of A<the P of D
and i got a negative value of K
2. (Original post by alesha98)
on the last question, how should I approach this?
I have found the coordinate of A and D and set up an equation so that:
the P of A<the P of D
and i got a negative value of K
Don't think that's going to help.

Well you know k is positive, so for a given value of P your profit line is going from top left to bottom right, in general.

For A to be the minimum value of P, you require it to be less than the value of B and C (and D).

Because the slope of P is negative, then the value of A, is automatically less than the value of C or D.

But it's not necessarily less than the value of B, unless you restrict k.
You may find it useful to draw a few lines through A with a range of k values, and it should then become clear what the lower limit on k is.

Now do a similar process with the maximum value of P.
3. (Original post by ghostwalker)
Don't think that's going to help.

Well you know k is positive, so for a given value of P your profit line is going from top left to bottom right, in general.

For A to be the minimum value of P, you require it to be less than the value of B and C (and D).

Because the slope of P is negative, then the value of A, is automatically less than the value of C or D.

But it's not necessarily less than the value of B, unless you restrict k.
You may find it useful to draw a few lines through A with a range of k values, and it should then become clear what the lower limit on k is.

Now do a similar process with the maximum value of P.
i understand that, but how am i able to do this in the exam? I found this question really hard and nonsense...
4. (Original post by ghostwalker)
Don't think that's going to help.

Well you know k is positive, so for a given value of P your profit line is going from top left to bottom right, in general.

For A to be the minimum value of P, you require it to be less than the value of B and C (and D).

Because the slope of P is negative, then the value of A, is automatically less than the value of C or D.

But it's not necessarily less than the value of B, unless you restrict k.
You may find it useful to draw a few lines through A with a range of k values, and it should then become clear what the lower limit on k is.

Now do a similar process with the maximum value of P.
also how about the maximum point, D? i need to compare A,B,C with D?
5. (Original post by alesha98)
i understand that, but how am i able to do this in the exam? I found this question really hard and nonsense...
Can't speak for an exam. For myself the moment you see a formula P=x+ky, where k varies, you know you've got the equation of a line with varying slope. You're told things about A and D, so imagine slapping the line through each of those two points in turn and varying the slope. That's what goes through my head.

(Original post by alesha98)
also how about the maximum point, D? i need to compare A,B,C with D?
You need to have a think about it. If you've understood the first bit, then try applying the same ideas, and see what you can come up with.

Although many people consider decision maths an easy option, IMO, it's not, and it needs a bit of creative thinking in a different fashion to core or mechanics.
6. (Original post by ghostwalker)
Can't speak for an exam. For myself the moment you see a formula P=x+ky, where k varies, you know you've got the equation of a line with varying slope. You're told things about A and D, so imagine slapping the line through each of those two points in turn and varying the slope. That's what goes through my head.

You need to have a think about it. If you've understood the first bit, then try applying the same ideas, and see what you can come up with.

Although many people consider decision maths an easy option, IMO, it's not, and it needs a bit of creative thinking in a different fashion to core or mechanics.
i dont think i am able to solve this question, i couldnt find any videos that explains it
7. This question took me a long time but I have figured it out. I looked at my graph and imagined what is the line with the least gradient that gives me A as a solution, I realised that it would be the line 4x+y=36 because when the line is moved across the page it gives me A and D as solutions as minimums and maximums. I then looked at the maximum gradient for a line and found 2x+y=36 worked.

You can then recognise that these equations if divided across are the same as P=x+ky, so I ended up with x + y/4 = 9 and x + y/2 = 18. I have found my k values of 1/4 and 1/2 but I also had to recognise that the lines I found had two minimum points so the inequality does not include these values, hence 1/4 < k < 1/2.

This method is purely graphical and I would recommend instead comparing the points A and B when they are subbed into P. You then say that the value of P must be less for A than B so you solve the inequality A < B. Then do the same with the points C and D with C < D.
8. (Original post by alesha98)
i dont think i am able to solve this question, i couldnt find any videos that explains it
For reference, here's the mark scheme - uses a different approach:

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