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    Hey guys, I've just got a question for the module: Decision 1.

    I've got this question on Linear Programming:
    And this is the graph/answer:
    http://prntscr.com/1567no


    This is part of the question:
    http://prntscr.com/1567rn

    The equation is P = 80x + 80y
    I just want to know how to actually plot that line accurately.

    Thanks.
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    Can anyone help?
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    You find an easy figure for P in this case we can use 8. So 8 = 80x+80y. Then use the cover up method t plot the line. Whatever figure you chose for P the line will always be the same gradient so wont affect your answer.


    Hope that helps
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    Hmm, so the gradient would be 1/10?
    How would we plot it?
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    Isn't it easier to rearrange for y = mx + P, then use y/x to find the gradient and plot on the graph?


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    I'm not sure how to do the gradient.

    The method I've been using was that if we set P to any value. I make 'x' equal to 0 to find where the line would be in the yaxis, then make y equal to 0 to find the point in the x axis. Then I can just connect it.

    But the problem is, if I pick a different number like in my first post's question P = 1600, so 1600 = 80x + 80y.
    By using the method I just used, x would be 20, so would y. But it wouldn't match the mark scheme's answer so I wouldn't get the marks.

    So I need help on this matter.
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    Your method for plotting the objective line is the same as mine. Even if you chose a different number eg. P= 2400 then x and y would still be equal. In the mark scheme you posted, the gradient of that objective line is the same as the gradient of your objective line, they just used P= 800 instead. In the edexcel mark schemes it just says "correct objective line plotted" and as far as I can tell, yours is correct. Hope I cleared things up
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    Hmm, yep they used 800.
    But what if I plotted it at x=20 And y=20 instead of being 10?

    although its same gradient, its not what the edexcel mark scheme shows.
 
 
 
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