Expert: Ahmed Salami - 5/8/2010

Question
QUESTION: Having these two conditions
xy=1
|x|<=|y|

How can be geometrically displayed the area or group of points,that will satisfy both conditions at the same time.
Thanking in advance.

ANSWER: Hi Domingos,
First you draw the graphs of y = 1/x and y = x. They intersect at the points (1,1) and (-1,-1).
I've added a picture showing the required area as hatched.
Actually, if you strictly meant xy = 1 then the region would only be the points on the y = 1/x curve.

NB: y = x is the green line and y = 1/x is the red line.

Regards

---------- FOLLOW-UP ----------

QUESTION: Dear Ahmed,
Thanks for your reply.
With it I got some hints of what should be or not be the requested area.
May be I am wrong, but according to your solution we can jump it false conclusions as:
take the following points that are in the areas you chose:
(4/3;3/4) in this case |4/3|<=|3/4| is false and not verifying the condition |x|<=|y|
(-2;0) again |-2|<=|0| is false not verifying the condition |x|<=|y|
So I believe that it is needed to draw also the line y = -x and then set the area requested as in the image I've uploaded.
Best regards and looking forward for your comments.

ANSWER: Hi Domingos,
Like i pointed out before, xy = 1 is not an inequality but an equation and so cannot represent a region. But assuming it to be xy < 1, i do get your point.
I have to agree with you though that there was some mixup in the hatching (although in a branch of simultaneous inequalities we tend to hatch the opposite direction). Anyway, its my mistake this time and, yes, we should add the line y = -x.
You've thought well but your graph is not without it's own fault too. I think you've misunderstood what happens to the curve xy = 1 when x < 0. I'm including a graph showing the required regions for both set of points separately and then combined and i'm sure it says everything.

I've always liked the careful readers and i'll say you've done well.

Regards

---------- FOLLOW-UP ----------

QUESTION: Dear Ahmed,
I agree with you about the mistake of using xy<1, when it should have remained as xy = 1.
To conclude from my side, I believe that the two initial conditions together should show two sections of the line y = 1/x, as I am displaying in a file that I upload now.
Many thanks for your work and kind words.
Anyhow I will wait for your final comments.
See you soon, I hope.Have a nice weekend.

Answer
Hello Domingos,
I should ask this, is there still anything you dispute about the solution given?
Assuming there is, lets put the maths aside for one second and instead just test points in the hatched (or unhatched) areas.
Consider the isolated graph showing the regions xy < 1. Is there something wrong with it? If there is, what is it?
Do the same for |x|≤|y|
If there's nothing wrong with both, then the third graph shows the complete solution. Notice that no point in the (-x & +y) quadrant satisfies the simultaneous condition.

Or maybe there's a point of yours that i'm missing. You can get back to me.

Regards