Question 1209092
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The difference of two positive numbers is 9. What is the minimum sum of their squares?

Absolute extrema/optimization
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<pre>
Let x be a greater of these two real numbers; y be a smaller of these two real numbers.

Then x - y = 9,  or  x = y+9.


They want you find the minimum of  {{{x^2 + y^2}}},  which is

    {{{(y+9)^2+ y^2}}} = {{{y^2 + 18y + 81 + y^2}}} = {{{2y^2 + 18y + 81}}}.


So, actually, we want to find the minimum of this quadratic function 

    f(y) = {{{2y^2 + 18y + 81}}}  on the set of positive real numbers  y > 0.


The global minimum of a quadratic function of the general form  f(y) = ay^2 + by + c  with positive
leading coefficient "a" is achieved at  {{{y[min]}}} = {{{-b/(2a)}}}.  In this case 

    {{{y[min]}}} = {{{-18/(2*2)}}} = -4.5,


which is negative value  <U>out of the scope of our consideration</U>.


In the area y > 0, we have the ascending branch of the parabola on the right from its minimum in negative
number. Had we consider the scope of non-negative real numbers y >= 0, then the minimum of the parabola 
f(y) = {{{2y^2 + 18y + 81}}}  would be at y = 0 with the value f(0) = 81.


But, according to the problem, our scope is the set of positive numbers y > 0, and in this domain, 
quadratic function f(y) = {{{2y^2 + 18y + 81}}} does not have the minimum, at all.


<U>ANSWER</U>.  In the set of positive numbers, the function  {{{x^2+y^2}}}  with the restriction  x-y = 9  has no minimum.
</pre>
Solved.



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Above, &nbsp;I told you the whole story with all details.


I could explain everything in more short terms, &nbsp;simply noticing that 
the parabola  &nbsp;f(y) = 2y^2 + 18y + 81  &nbsp;has two positive terms  &nbsp;2y^2  &nbsp;and  &nbsp;18y
in the domain &nbsp;y > 0.


So, &nbsp;the values of &nbsp;f(y) &nbsp;go up as &nbsp;"y" &nbsp;goes to the right from zero, 
which means that function &nbsp;f(y) &nbsp;raises there.


In opposite, &nbsp;function &nbsp;f(y) &nbsp;goes down as &nbsp;"y" approaches &nbsp;0 &nbsp;from the right side.
But &nbsp;"y" &nbsp;never gets &nbsp;0 &nbsp;in the positive domain - so we never get the minimum of &nbsp;f(y) 
in the positive domain &nbsp;y > 0.


With these two explanations, &nbsp;you have now a complete picture and a complete vision to the problem.