ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE General & Further Mathematics => Topic started by: k31453 on November 01, 2012, 08:13:37 pm
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How do you do this question ?
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Well. If we try the first alternative we get x < a -y.
We can re-arrange this for x-a < -y.
Which is the same as y < a - x.
Now suppose a was a number, like 5.
y < 5 - x has a y -intercept at (0,5) so (0,a)
And x -intercept (5,0) so (a,0)
Also, substitute in 0,0 to determine where to shade.
0 < 5 is true, so shade away from the origin. (i.e. 0 < a).
This confirms our choice of the alternative A, as it is a dotted line going through the points (0,a) and (a,0) and is shaded on the opposite side to the origin. I hope this helped :)
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You could do the same thing for E as well though, couldnt you?
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You could do the same thing for E as well though, couldnt you?
Almost, but the last step of finding the feasible region would be different.
You can re-arrange option E to say y < x - b. Which is an equation on the graph.
But, if you substitute in the point (0,0) you get 0 < - b. Which is false, as 0 is > -b.
So that means you would shade in the side including the origin, which is not what they have done in the question.
So close, but no cigar unfortunately :)
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Hi!
Assuming that the feasible region is the unshaded white area, the answer would be E? Or am I mistaken?
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ohk got it :)