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# Decision 1 - Plotting the Objective Line Properly. Watch

1. 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.
2. Can anyone help?
3. 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
4. Hmm, so the gradient would be 1/10?
How would we plot it?
5. 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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6. 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.
7. 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
8. 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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