Question 1148538
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            This problem is  VERY  SIMPLE.


            Moreover, it is  EXTREMELY  SIMPLE,  and I will show it to you right now.



<pre>
Our function is  {{{(x+y)/x}}} = 1 + {{{y/x}}}.      (1)


The given area is the square  ABCD  in quadrant  QII  with the vertices  

A= (-4,2),  B= (-2,2),  C= (-2,4),  D= (-4,4).



The given function, OBVIOUSLY, adds the ratio  {{{y/x}}}  to the value of 1.


But this ratio  {{{y/x}}}  is negative in QII.


Thus the function (1), actually, adds NEGATIVE amount to 1.



    - When the function (1) will have the largest value ?

    - OBVIOUSLY, when this negative addend  {{{y/x}}}  will have minimal absolute value.



        - When this addend  {{{y/x}}}  will have minimal absolute value ?

        - OBVIOUSLY,  when positive "y" is minimal and negative "x" has maximal absolute value.



             - Now, what is the point of the given square, where positive "y" is minimal and negative "x" has maximal absolute value ?

             - OBVIOUSLY, this point is  A= (-4,2).



Then the larges possible value of  the function  {{{(x+y)/x}}}   is   {{{1 + y/x}}} = {{{1 + (2/(-4))}}} = 1 - {{{1/2}}} = {{{1/2}}}.    <U>ANSWER</U>
</pre>

Solved, explained, answered and completed.




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O-o-o, tutor @greenestamps successfully retold my solution to this problem !


Congratulations (!) (!)


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