Maximum value of $x$ when equality is given











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$$ x + y = sqrt{x} + sqrt{y} $$
Find maximum value of $x$. $x$ and $y$ are reals.










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  • "Find maximum value...of $x$"? What does this mean?
    – DonAntonio
    Nov 17 at 20:17












  • it means most probably to find max value of x satisfying the equation
    – maveric
    Nov 17 at 20:18






  • 3




    Ok, I think I understand now. When $;x,yinBbb N;$ (since they obviously must be non-negative), what is the maximum value for some $;x;$ fulfilling that no matter what $;y;$ we take...A little odd question.
    – DonAntonio
    Nov 17 at 20:20










  • The question makes no much sense, as @DonAntonio suggests.
    – Rebellos
    Nov 17 at 20:22










  • Solve as quadric with respect to $sqrt{x}$ and then maximize the root value + the condition on integers. Would that not work?
    – Makina
    Nov 17 at 20:23















up vote
4
down vote

favorite












$$ x + y = sqrt{x} + sqrt{y} $$
Find maximum value of $x$. $x$ and $y$ are reals.










share|cite|improve this question
























  • "Find maximum value...of $x$"? What does this mean?
    – DonAntonio
    Nov 17 at 20:17












  • it means most probably to find max value of x satisfying the equation
    – maveric
    Nov 17 at 20:18






  • 3




    Ok, I think I understand now. When $;x,yinBbb N;$ (since they obviously must be non-negative), what is the maximum value for some $;x;$ fulfilling that no matter what $;y;$ we take...A little odd question.
    – DonAntonio
    Nov 17 at 20:20










  • The question makes no much sense, as @DonAntonio suggests.
    – Rebellos
    Nov 17 at 20:22










  • Solve as quadric with respect to $sqrt{x}$ and then maximize the root value + the condition on integers. Would that not work?
    – Makina
    Nov 17 at 20:23













up vote
4
down vote

favorite









up vote
4
down vote

favorite











$$ x + y = sqrt{x} + sqrt{y} $$
Find maximum value of $x$. $x$ and $y$ are reals.










share|cite|improve this question















$$ x + y = sqrt{x} + sqrt{y} $$
Find maximum value of $x$. $x$ and $y$ are reals.







algebra-precalculus






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edited Nov 18 at 8:00









iBug

1427




1427










asked Nov 17 at 20:13









maveric

61611




61611












  • "Find maximum value...of $x$"? What does this mean?
    – DonAntonio
    Nov 17 at 20:17












  • it means most probably to find max value of x satisfying the equation
    – maveric
    Nov 17 at 20:18






  • 3




    Ok, I think I understand now. When $;x,yinBbb N;$ (since they obviously must be non-negative), what is the maximum value for some $;x;$ fulfilling that no matter what $;y;$ we take...A little odd question.
    – DonAntonio
    Nov 17 at 20:20










  • The question makes no much sense, as @DonAntonio suggests.
    – Rebellos
    Nov 17 at 20:22










  • Solve as quadric with respect to $sqrt{x}$ and then maximize the root value + the condition on integers. Would that not work?
    – Makina
    Nov 17 at 20:23


















  • "Find maximum value...of $x$"? What does this mean?
    – DonAntonio
    Nov 17 at 20:17












  • it means most probably to find max value of x satisfying the equation
    – maveric
    Nov 17 at 20:18






  • 3




    Ok, I think I understand now. When $;x,yinBbb N;$ (since they obviously must be non-negative), what is the maximum value for some $;x;$ fulfilling that no matter what $;y;$ we take...A little odd question.
    – DonAntonio
    Nov 17 at 20:20










  • The question makes no much sense, as @DonAntonio suggests.
    – Rebellos
    Nov 17 at 20:22










  • Solve as quadric with respect to $sqrt{x}$ and then maximize the root value + the condition on integers. Would that not work?
    – Makina
    Nov 17 at 20:23
















"Find maximum value...of $x$"? What does this mean?
– DonAntonio
Nov 17 at 20:17






"Find maximum value...of $x$"? What does this mean?
– DonAntonio
Nov 17 at 20:17














it means most probably to find max value of x satisfying the equation
– maveric
Nov 17 at 20:18




it means most probably to find max value of x satisfying the equation
– maveric
Nov 17 at 20:18




3




3




Ok, I think I understand now. When $;x,yinBbb N;$ (since they obviously must be non-negative), what is the maximum value for some $;x;$ fulfilling that no matter what $;y;$ we take...A little odd question.
– DonAntonio
Nov 17 at 20:20




Ok, I think I understand now. When $;x,yinBbb N;$ (since they obviously must be non-negative), what is the maximum value for some $;x;$ fulfilling that no matter what $;y;$ we take...A little odd question.
– DonAntonio
Nov 17 at 20:20












The question makes no much sense, as @DonAntonio suggests.
– Rebellos
Nov 17 at 20:22




The question makes no much sense, as @DonAntonio suggests.
– Rebellos
Nov 17 at 20:22












Solve as quadric with respect to $sqrt{x}$ and then maximize the root value + the condition on integers. Would that not work?
– Makina
Nov 17 at 20:23




Solve as quadric with respect to $sqrt{x}$ and then maximize the root value + the condition on integers. Would that not work?
– Makina
Nov 17 at 20:23










7 Answers
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4
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Try this method of completing the square.



Let $a=sqrt x$ and $b=sqrt y$ so that $a^2+b^2=a+b$ and $4a^2-4a+1+4b^2-4b+1=2$ ie $(2a-1)^2+(2b-1)^2=2$






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    Obviously $x,yge0$. We can therefore write $x=r^2cos^2theta$ and $y=r^2sin^2theta$. Then the equation becomes $$r^2(cos^2theta+sin^2theta)=r(costheta+sintheta)$$
    with $theta in [0,fracpi2]$ and $rgt 0$. Dividing by $r$ and multiplying by $costheta$ we get $$sqrt x=(costheta+sintheta)costheta=cos^2theta+sinthetacostheta$$
    Now find the maximum of the expression on the right. Use formula for the sine and cosine for double angle.






    share|cite|improve this answer




























      up vote
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      enter image description here



      If you set $f(x)=x-sqrt{x}$ then you want to solve $f(x)=-f(y)$



      The maximum $x$ is reached for the line in green intersecting both curves $f$ and $-f$ at the maximal possible altitude.



      So we have to solve $f'(y)=0iff1-dfrac1{2sqrt{y}}=0iff y=frac 14$



      The corresponding $x$ verifies $f(x)=-f(y)=dfrac 14$ and this is just a quadratic polynomial to solve to find the $x$ value.



      The maximum is then $x=frac{3+2sqrt{2}}4approx 1.4571$






      share|cite|improve this answer



















      • 1




        dont understant what you trying to convey?
        – maveric
        Nov 17 at 20:50


















      up vote
      3
      down vote













      Since $x+ y = sqrt{x} + sqrt{y}$ is symmetric, WLOG, we can consider $xge yge 0$ or $y=ax, 0le ale 1$. Then:
      $$x+ ax = sqrt{x} + sqrt{ax} Rightarrow color{red}{sqrt{x}}=frac{1+sqrt{a}}{1+a} to text{max s.t. $0le ale 1$}\
      f'(a)=left(frac{1+sqrt{a}}{1+a}right)'=0 Rightarrow frac{frac1{2sqrt{a}}(1+a)-(1+sqrt{a})}{(1+a)^2}=0 Rightarrow \
      a+2sqrt{a}-1=0 Rightarrow a=3-2sqrt{2}approx 0.17.$$

      Note that the function $f(a)$ is continuous at $0le ale 1$ and $f(0)=f(1)=1$ and $f(3-2sqrt{2})=frac12(1+sqrt{2})approx 1.207$. Hence it must be the maximum point (see Desmos graph).



      So, the maximum value of $x$ is:
      $$color{red}x=left(frac{1+sqrt{3-2sqrt{2}}}{1+(3-2sqrt{2})}right)^2=frac14left(1+sqrt{2}right)^2approx 1.457.$$






      share|cite|improve this answer























      • I don't think that's right. See zwim's answer, or just graph the curve in question to see that the right value should be around $1.46$.
        – YiFan
        Nov 18 at 6:25






      • 2




        $1.207^2approx 1.457$ which is the right answer, so the method is fine. There is an error in the last calculation of $x$.
        – zwim
        Nov 18 at 6:32






      • 1




        @zwim, thank you for pointing to the error. Fixed it.
        – farruhota
        Nov 18 at 6:43










      • +1. Nice method.
        – Taladris
        Nov 19 at 0:45


















      up vote
      1
      down vote













      Observe that



      $$x+y=sqrt x+sqrt yiff sqrt xleft(sqrt x-1right)=sqrt yleft(1-sqrt yright)$$



      Now, for any values $;x,y>1;$ , the left side above will be positive whereas the right side will be negative (can you see why?), thus we must restrict the possible natural values...Can you take it from here?






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      • sorry the question is edited. it was real.
        – maveric
        Nov 17 at 20:25










      • can we say then 1?
        – maveric
        Nov 17 at 20:25










      • The above solves completely the first, original question. Now that it has changed to "real" it at least can help you to realize that $;y;$ cannot be greater than $;1;$ and $;x;$ smaller than one, for example...
        – DonAntonio
        Nov 17 at 20:39


















      up vote
      0
      down vote













      Let $f(x)=x-sqrt x$. The question becomes trying to find two real values $x,y$ so that $f(x)=-f(y)$, then maximising one of these values without regard for what happens to the other. After a bit of thinking analysing derivatives, it becomes clear that we want to minimise $f(x)$ so that $f(y)$, and hence $y$, can be maximised. So just set $f'(x)=0$ to find the local minimum value to be $-1/4$, and hence the maximum value of $y$ satisfies $f(y)=1/4$. This is a simple quadratic equation, solve, and you're done!






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        There are some caveats (we need to assume the existence of a maximum for example) but Lagrange's multipliers method work well here to find the value of the maximum:



        Let $f(x,y)=x$ and $g(x,y)=x-sqrt{x}+y-sqrt{y}$. We want to find the maximum value of $f$ under constraint $g(x,y)=0$.



        We have $nabla f=langle 1,0 rangle$ and $nabla g = langle 1-frac{1}{2sqrt{x}},1-frac{1}{2sqrt{y}} rangle$. At a maximum, we have $nabla g=lambda nabla f$ for some real number $lambda$. This implies that $1-frac{1}{2sqrt{y}}=0$, that is $y=frac{1}{4}$.



        Using the constraint, we have $x-sqrt{x}-frac{1}{4}=0$ which is equivalent to $a^2-a-frac{1}{4}$ and $sqrt{x}=ageqslant 0$. We obtain $a=frac{1+sqrt{2}}{2}$, so $x=frac{(1+sqrt{2})^2}{4}=frac{3+2sqrt{2}}{4}$.





        (sloppy) proof of the existence of the global maximum value: The constraint $x+y=sqrt{x}+sqrt{y}$ implies that $xgeqslant 0$ and $ygeqslant 0$. Moreover, if $x$ or $y$ is large enough (x>R or y>R for some $R$ that I am too lazy to determine explicitly :p), $x+y>sqrt{x}+sqrt{y}$, so we can assume that $x,y in [0,R]$.



        Now, the function $f(x,y)=x$ is continuous and the set $S=left{ (x,y)in [0,R]^2; middle|; g(x,y)=0; right}$ is compact (closed and bounded in ${mathbb R}^2$) so, by the Extreme Value Theorem, $f$ attains its global extreme values on $S$. The global minimum value is obviously obtained at $(0,0)$, while the other one is located by Lagrange's method.






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          7 Answers
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          7 Answers
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          up vote
          4
          down vote













          Try this method of completing the square.



          Let $a=sqrt x$ and $b=sqrt y$ so that $a^2+b^2=a+b$ and $4a^2-4a+1+4b^2-4b+1=2$ ie $(2a-1)^2+(2b-1)^2=2$






          share|cite|improve this answer

























            up vote
            4
            down vote













            Try this method of completing the square.



            Let $a=sqrt x$ and $b=sqrt y$ so that $a^2+b^2=a+b$ and $4a^2-4a+1+4b^2-4b+1=2$ ie $(2a-1)^2+(2b-1)^2=2$






            share|cite|improve this answer























              up vote
              4
              down vote










              up vote
              4
              down vote









              Try this method of completing the square.



              Let $a=sqrt x$ and $b=sqrt y$ so that $a^2+b^2=a+b$ and $4a^2-4a+1+4b^2-4b+1=2$ ie $(2a-1)^2+(2b-1)^2=2$






              share|cite|improve this answer












              Try this method of completing the square.



              Let $a=sqrt x$ and $b=sqrt y$ so that $a^2+b^2=a+b$ and $4a^2-4a+1+4b^2-4b+1=2$ ie $(2a-1)^2+(2b-1)^2=2$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 17 at 20:25









              Mark Bennet

              79.9k980178




              79.9k980178






















                  up vote
                  3
                  down vote













                  Obviously $x,yge0$. We can therefore write $x=r^2cos^2theta$ and $y=r^2sin^2theta$. Then the equation becomes $$r^2(cos^2theta+sin^2theta)=r(costheta+sintheta)$$
                  with $theta in [0,fracpi2]$ and $rgt 0$. Dividing by $r$ and multiplying by $costheta$ we get $$sqrt x=(costheta+sintheta)costheta=cos^2theta+sinthetacostheta$$
                  Now find the maximum of the expression on the right. Use formula for the sine and cosine for double angle.






                  share|cite|improve this answer

























                    up vote
                    3
                    down vote













                    Obviously $x,yge0$. We can therefore write $x=r^2cos^2theta$ and $y=r^2sin^2theta$. Then the equation becomes $$r^2(cos^2theta+sin^2theta)=r(costheta+sintheta)$$
                    with $theta in [0,fracpi2]$ and $rgt 0$. Dividing by $r$ and multiplying by $costheta$ we get $$sqrt x=(costheta+sintheta)costheta=cos^2theta+sinthetacostheta$$
                    Now find the maximum of the expression on the right. Use formula for the sine and cosine for double angle.






                    share|cite|improve this answer























                      up vote
                      3
                      down vote










                      up vote
                      3
                      down vote









                      Obviously $x,yge0$. We can therefore write $x=r^2cos^2theta$ and $y=r^2sin^2theta$. Then the equation becomes $$r^2(cos^2theta+sin^2theta)=r(costheta+sintheta)$$
                      with $theta in [0,fracpi2]$ and $rgt 0$. Dividing by $r$ and multiplying by $costheta$ we get $$sqrt x=(costheta+sintheta)costheta=cos^2theta+sinthetacostheta$$
                      Now find the maximum of the expression on the right. Use formula for the sine and cosine for double angle.






                      share|cite|improve this answer












                      Obviously $x,yge0$. We can therefore write $x=r^2cos^2theta$ and $y=r^2sin^2theta$. Then the equation becomes $$r^2(cos^2theta+sin^2theta)=r(costheta+sintheta)$$
                      with $theta in [0,fracpi2]$ and $rgt 0$. Dividing by $r$ and multiplying by $costheta$ we get $$sqrt x=(costheta+sintheta)costheta=cos^2theta+sinthetacostheta$$
                      Now find the maximum of the expression on the right. Use formula for the sine and cosine for double angle.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 17 at 20:47









                      Andrei

                      10.3k21025




                      10.3k21025






















                          up vote
                          3
                          down vote













                          enter image description here



                          If you set $f(x)=x-sqrt{x}$ then you want to solve $f(x)=-f(y)$



                          The maximum $x$ is reached for the line in green intersecting both curves $f$ and $-f$ at the maximal possible altitude.



                          So we have to solve $f'(y)=0iff1-dfrac1{2sqrt{y}}=0iff y=frac 14$



                          The corresponding $x$ verifies $f(x)=-f(y)=dfrac 14$ and this is just a quadratic polynomial to solve to find the $x$ value.



                          The maximum is then $x=frac{3+2sqrt{2}}4approx 1.4571$






                          share|cite|improve this answer



















                          • 1




                            dont understant what you trying to convey?
                            – maveric
                            Nov 17 at 20:50















                          up vote
                          3
                          down vote













                          enter image description here



                          If you set $f(x)=x-sqrt{x}$ then you want to solve $f(x)=-f(y)$



                          The maximum $x$ is reached for the line in green intersecting both curves $f$ and $-f$ at the maximal possible altitude.



                          So we have to solve $f'(y)=0iff1-dfrac1{2sqrt{y}}=0iff y=frac 14$



                          The corresponding $x$ verifies $f(x)=-f(y)=dfrac 14$ and this is just a quadratic polynomial to solve to find the $x$ value.



                          The maximum is then $x=frac{3+2sqrt{2}}4approx 1.4571$






                          share|cite|improve this answer



















                          • 1




                            dont understant what you trying to convey?
                            – maveric
                            Nov 17 at 20:50













                          up vote
                          3
                          down vote










                          up vote
                          3
                          down vote









                          enter image description here



                          If you set $f(x)=x-sqrt{x}$ then you want to solve $f(x)=-f(y)$



                          The maximum $x$ is reached for the line in green intersecting both curves $f$ and $-f$ at the maximal possible altitude.



                          So we have to solve $f'(y)=0iff1-dfrac1{2sqrt{y}}=0iff y=frac 14$



                          The corresponding $x$ verifies $f(x)=-f(y)=dfrac 14$ and this is just a quadratic polynomial to solve to find the $x$ value.



                          The maximum is then $x=frac{3+2sqrt{2}}4approx 1.4571$






                          share|cite|improve this answer














                          enter image description here



                          If you set $f(x)=x-sqrt{x}$ then you want to solve $f(x)=-f(y)$



                          The maximum $x$ is reached for the line in green intersecting both curves $f$ and $-f$ at the maximal possible altitude.



                          So we have to solve $f'(y)=0iff1-dfrac1{2sqrt{y}}=0iff y=frac 14$



                          The corresponding $x$ verifies $f(x)=-f(y)=dfrac 14$ and this is just a quadratic polynomial to solve to find the $x$ value.



                          The maximum is then $x=frac{3+2sqrt{2}}4approx 1.4571$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Nov 18 at 6:21

























                          answered Nov 17 at 20:47









                          zwim

                          11.2k628




                          11.2k628








                          • 1




                            dont understant what you trying to convey?
                            – maveric
                            Nov 17 at 20:50














                          • 1




                            dont understant what you trying to convey?
                            – maveric
                            Nov 17 at 20:50








                          1




                          1




                          dont understant what you trying to convey?
                          – maveric
                          Nov 17 at 20:50




                          dont understant what you trying to convey?
                          – maveric
                          Nov 17 at 20:50










                          up vote
                          3
                          down vote













                          Since $x+ y = sqrt{x} + sqrt{y}$ is symmetric, WLOG, we can consider $xge yge 0$ or $y=ax, 0le ale 1$. Then:
                          $$x+ ax = sqrt{x} + sqrt{ax} Rightarrow color{red}{sqrt{x}}=frac{1+sqrt{a}}{1+a} to text{max s.t. $0le ale 1$}\
                          f'(a)=left(frac{1+sqrt{a}}{1+a}right)'=0 Rightarrow frac{frac1{2sqrt{a}}(1+a)-(1+sqrt{a})}{(1+a)^2}=0 Rightarrow \
                          a+2sqrt{a}-1=0 Rightarrow a=3-2sqrt{2}approx 0.17.$$

                          Note that the function $f(a)$ is continuous at $0le ale 1$ and $f(0)=f(1)=1$ and $f(3-2sqrt{2})=frac12(1+sqrt{2})approx 1.207$. Hence it must be the maximum point (see Desmos graph).



                          So, the maximum value of $x$ is:
                          $$color{red}x=left(frac{1+sqrt{3-2sqrt{2}}}{1+(3-2sqrt{2})}right)^2=frac14left(1+sqrt{2}right)^2approx 1.457.$$






                          share|cite|improve this answer























                          • I don't think that's right. See zwim's answer, or just graph the curve in question to see that the right value should be around $1.46$.
                            – YiFan
                            Nov 18 at 6:25






                          • 2




                            $1.207^2approx 1.457$ which is the right answer, so the method is fine. There is an error in the last calculation of $x$.
                            – zwim
                            Nov 18 at 6:32






                          • 1




                            @zwim, thank you for pointing to the error. Fixed it.
                            – farruhota
                            Nov 18 at 6:43










                          • +1. Nice method.
                            – Taladris
                            Nov 19 at 0:45















                          up vote
                          3
                          down vote













                          Since $x+ y = sqrt{x} + sqrt{y}$ is symmetric, WLOG, we can consider $xge yge 0$ or $y=ax, 0le ale 1$. Then:
                          $$x+ ax = sqrt{x} + sqrt{ax} Rightarrow color{red}{sqrt{x}}=frac{1+sqrt{a}}{1+a} to text{max s.t. $0le ale 1$}\
                          f'(a)=left(frac{1+sqrt{a}}{1+a}right)'=0 Rightarrow frac{frac1{2sqrt{a}}(1+a)-(1+sqrt{a})}{(1+a)^2}=0 Rightarrow \
                          a+2sqrt{a}-1=0 Rightarrow a=3-2sqrt{2}approx 0.17.$$

                          Note that the function $f(a)$ is continuous at $0le ale 1$ and $f(0)=f(1)=1$ and $f(3-2sqrt{2})=frac12(1+sqrt{2})approx 1.207$. Hence it must be the maximum point (see Desmos graph).



                          So, the maximum value of $x$ is:
                          $$color{red}x=left(frac{1+sqrt{3-2sqrt{2}}}{1+(3-2sqrt{2})}right)^2=frac14left(1+sqrt{2}right)^2approx 1.457.$$






                          share|cite|improve this answer























                          • I don't think that's right. See zwim's answer, or just graph the curve in question to see that the right value should be around $1.46$.
                            – YiFan
                            Nov 18 at 6:25






                          • 2




                            $1.207^2approx 1.457$ which is the right answer, so the method is fine. There is an error in the last calculation of $x$.
                            – zwim
                            Nov 18 at 6:32






                          • 1




                            @zwim, thank you for pointing to the error. Fixed it.
                            – farruhota
                            Nov 18 at 6:43










                          • +1. Nice method.
                            – Taladris
                            Nov 19 at 0:45













                          up vote
                          3
                          down vote










                          up vote
                          3
                          down vote









                          Since $x+ y = sqrt{x} + sqrt{y}$ is symmetric, WLOG, we can consider $xge yge 0$ or $y=ax, 0le ale 1$. Then:
                          $$x+ ax = sqrt{x} + sqrt{ax} Rightarrow color{red}{sqrt{x}}=frac{1+sqrt{a}}{1+a} to text{max s.t. $0le ale 1$}\
                          f'(a)=left(frac{1+sqrt{a}}{1+a}right)'=0 Rightarrow frac{frac1{2sqrt{a}}(1+a)-(1+sqrt{a})}{(1+a)^2}=0 Rightarrow \
                          a+2sqrt{a}-1=0 Rightarrow a=3-2sqrt{2}approx 0.17.$$

                          Note that the function $f(a)$ is continuous at $0le ale 1$ and $f(0)=f(1)=1$ and $f(3-2sqrt{2})=frac12(1+sqrt{2})approx 1.207$. Hence it must be the maximum point (see Desmos graph).



                          So, the maximum value of $x$ is:
                          $$color{red}x=left(frac{1+sqrt{3-2sqrt{2}}}{1+(3-2sqrt{2})}right)^2=frac14left(1+sqrt{2}right)^2approx 1.457.$$






                          share|cite|improve this answer














                          Since $x+ y = sqrt{x} + sqrt{y}$ is symmetric, WLOG, we can consider $xge yge 0$ or $y=ax, 0le ale 1$. Then:
                          $$x+ ax = sqrt{x} + sqrt{ax} Rightarrow color{red}{sqrt{x}}=frac{1+sqrt{a}}{1+a} to text{max s.t. $0le ale 1$}\
                          f'(a)=left(frac{1+sqrt{a}}{1+a}right)'=0 Rightarrow frac{frac1{2sqrt{a}}(1+a)-(1+sqrt{a})}{(1+a)^2}=0 Rightarrow \
                          a+2sqrt{a}-1=0 Rightarrow a=3-2sqrt{2}approx 0.17.$$

                          Note that the function $f(a)$ is continuous at $0le ale 1$ and $f(0)=f(1)=1$ and $f(3-2sqrt{2})=frac12(1+sqrt{2})approx 1.207$. Hence it must be the maximum point (see Desmos graph).



                          So, the maximum value of $x$ is:
                          $$color{red}x=left(frac{1+sqrt{3-2sqrt{2}}}{1+(3-2sqrt{2})}right)^2=frac14left(1+sqrt{2}right)^2approx 1.457.$$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Nov 18 at 6:39

























                          answered Nov 18 at 5:32









                          farruhota

                          18k2736




                          18k2736












                          • I don't think that's right. See zwim's answer, or just graph the curve in question to see that the right value should be around $1.46$.
                            – YiFan
                            Nov 18 at 6:25






                          • 2




                            $1.207^2approx 1.457$ which is the right answer, so the method is fine. There is an error in the last calculation of $x$.
                            – zwim
                            Nov 18 at 6:32






                          • 1




                            @zwim, thank you for pointing to the error. Fixed it.
                            – farruhota
                            Nov 18 at 6:43










                          • +1. Nice method.
                            – Taladris
                            Nov 19 at 0:45


















                          • I don't think that's right. See zwim's answer, or just graph the curve in question to see that the right value should be around $1.46$.
                            – YiFan
                            Nov 18 at 6:25






                          • 2




                            $1.207^2approx 1.457$ which is the right answer, so the method is fine. There is an error in the last calculation of $x$.
                            – zwim
                            Nov 18 at 6:32






                          • 1




                            @zwim, thank you for pointing to the error. Fixed it.
                            – farruhota
                            Nov 18 at 6:43










                          • +1. Nice method.
                            – Taladris
                            Nov 19 at 0:45
















                          I don't think that's right. See zwim's answer, or just graph the curve in question to see that the right value should be around $1.46$.
                          – YiFan
                          Nov 18 at 6:25




                          I don't think that's right. See zwim's answer, or just graph the curve in question to see that the right value should be around $1.46$.
                          – YiFan
                          Nov 18 at 6:25




                          2




                          2




                          $1.207^2approx 1.457$ which is the right answer, so the method is fine. There is an error in the last calculation of $x$.
                          – zwim
                          Nov 18 at 6:32




                          $1.207^2approx 1.457$ which is the right answer, so the method is fine. There is an error in the last calculation of $x$.
                          – zwim
                          Nov 18 at 6:32




                          1




                          1




                          @zwim, thank you for pointing to the error. Fixed it.
                          – farruhota
                          Nov 18 at 6:43




                          @zwim, thank you for pointing to the error. Fixed it.
                          – farruhota
                          Nov 18 at 6:43












                          +1. Nice method.
                          – Taladris
                          Nov 19 at 0:45




                          +1. Nice method.
                          – Taladris
                          Nov 19 at 0:45










                          up vote
                          1
                          down vote













                          Observe that



                          $$x+y=sqrt x+sqrt yiff sqrt xleft(sqrt x-1right)=sqrt yleft(1-sqrt yright)$$



                          Now, for any values $;x,y>1;$ , the left side above will be positive whereas the right side will be negative (can you see why?), thus we must restrict the possible natural values...Can you take it from here?






                          share|cite|improve this answer





















                          • sorry the question is edited. it was real.
                            – maveric
                            Nov 17 at 20:25










                          • can we say then 1?
                            – maveric
                            Nov 17 at 20:25










                          • The above solves completely the first, original question. Now that it has changed to "real" it at least can help you to realize that $;y;$ cannot be greater than $;1;$ and $;x;$ smaller than one, for example...
                            – DonAntonio
                            Nov 17 at 20:39















                          up vote
                          1
                          down vote













                          Observe that



                          $$x+y=sqrt x+sqrt yiff sqrt xleft(sqrt x-1right)=sqrt yleft(1-sqrt yright)$$



                          Now, for any values $;x,y>1;$ , the left side above will be positive whereas the right side will be negative (can you see why?), thus we must restrict the possible natural values...Can you take it from here?






                          share|cite|improve this answer





















                          • sorry the question is edited. it was real.
                            – maveric
                            Nov 17 at 20:25










                          • can we say then 1?
                            – maveric
                            Nov 17 at 20:25










                          • The above solves completely the first, original question. Now that it has changed to "real" it at least can help you to realize that $;y;$ cannot be greater than $;1;$ and $;x;$ smaller than one, for example...
                            – DonAntonio
                            Nov 17 at 20:39













                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          Observe that



                          $$x+y=sqrt x+sqrt yiff sqrt xleft(sqrt x-1right)=sqrt yleft(1-sqrt yright)$$



                          Now, for any values $;x,y>1;$ , the left side above will be positive whereas the right side will be negative (can you see why?), thus we must restrict the possible natural values...Can you take it from here?






                          share|cite|improve this answer












                          Observe that



                          $$x+y=sqrt x+sqrt yiff sqrt xleft(sqrt x-1right)=sqrt yleft(1-sqrt yright)$$



                          Now, for any values $;x,y>1;$ , the left side above will be positive whereas the right side will be negative (can you see why?), thus we must restrict the possible natural values...Can you take it from here?







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 17 at 20:24









                          DonAntonio

                          176k1491224




                          176k1491224












                          • sorry the question is edited. it was real.
                            – maveric
                            Nov 17 at 20:25










                          • can we say then 1?
                            – maveric
                            Nov 17 at 20:25










                          • The above solves completely the first, original question. Now that it has changed to "real" it at least can help you to realize that $;y;$ cannot be greater than $;1;$ and $;x;$ smaller than one, for example...
                            – DonAntonio
                            Nov 17 at 20:39


















                          • sorry the question is edited. it was real.
                            – maveric
                            Nov 17 at 20:25










                          • can we say then 1?
                            – maveric
                            Nov 17 at 20:25










                          • The above solves completely the first, original question. Now that it has changed to "real" it at least can help you to realize that $;y;$ cannot be greater than $;1;$ and $;x;$ smaller than one, for example...
                            – DonAntonio
                            Nov 17 at 20:39
















                          sorry the question is edited. it was real.
                          – maveric
                          Nov 17 at 20:25




                          sorry the question is edited. it was real.
                          – maveric
                          Nov 17 at 20:25












                          can we say then 1?
                          – maveric
                          Nov 17 at 20:25




                          can we say then 1?
                          – maveric
                          Nov 17 at 20:25












                          The above solves completely the first, original question. Now that it has changed to "real" it at least can help you to realize that $;y;$ cannot be greater than $;1;$ and $;x;$ smaller than one, for example...
                          – DonAntonio
                          Nov 17 at 20:39




                          The above solves completely the first, original question. Now that it has changed to "real" it at least can help you to realize that $;y;$ cannot be greater than $;1;$ and $;x;$ smaller than one, for example...
                          – DonAntonio
                          Nov 17 at 20:39










                          up vote
                          0
                          down vote













                          Let $f(x)=x-sqrt x$. The question becomes trying to find two real values $x,y$ so that $f(x)=-f(y)$, then maximising one of these values without regard for what happens to the other. After a bit of thinking analysing derivatives, it becomes clear that we want to minimise $f(x)$ so that $f(y)$, and hence $y$, can be maximised. So just set $f'(x)=0$ to find the local minimum value to be $-1/4$, and hence the maximum value of $y$ satisfies $f(y)=1/4$. This is a simple quadratic equation, solve, and you're done!






                          share|cite|improve this answer

























                            up vote
                            0
                            down vote













                            Let $f(x)=x-sqrt x$. The question becomes trying to find two real values $x,y$ so that $f(x)=-f(y)$, then maximising one of these values without regard for what happens to the other. After a bit of thinking analysing derivatives, it becomes clear that we want to minimise $f(x)$ so that $f(y)$, and hence $y$, can be maximised. So just set $f'(x)=0$ to find the local minimum value to be $-1/4$, and hence the maximum value of $y$ satisfies $f(y)=1/4$. This is a simple quadratic equation, solve, and you're done!






                            share|cite|improve this answer























                              up vote
                              0
                              down vote










                              up vote
                              0
                              down vote









                              Let $f(x)=x-sqrt x$. The question becomes trying to find two real values $x,y$ so that $f(x)=-f(y)$, then maximising one of these values without regard for what happens to the other. After a bit of thinking analysing derivatives, it becomes clear that we want to minimise $f(x)$ so that $f(y)$, and hence $y$, can be maximised. So just set $f'(x)=0$ to find the local minimum value to be $-1/4$, and hence the maximum value of $y$ satisfies $f(y)=1/4$. This is a simple quadratic equation, solve, and you're done!






                              share|cite|improve this answer












                              Let $f(x)=x-sqrt x$. The question becomes trying to find two real values $x,y$ so that $f(x)=-f(y)$, then maximising one of these values without regard for what happens to the other. After a bit of thinking analysing derivatives, it becomes clear that we want to minimise $f(x)$ so that $f(y)$, and hence $y$, can be maximised. So just set $f'(x)=0$ to find the local minimum value to be $-1/4$, and hence the maximum value of $y$ satisfies $f(y)=1/4$. This is a simple quadratic equation, solve, and you're done!







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Nov 18 at 6:22









                              YiFan

                              1,7441315




                              1,7441315






















                                  up vote
                                  0
                                  down vote













                                  There are some caveats (we need to assume the existence of a maximum for example) but Lagrange's multipliers method work well here to find the value of the maximum:



                                  Let $f(x,y)=x$ and $g(x,y)=x-sqrt{x}+y-sqrt{y}$. We want to find the maximum value of $f$ under constraint $g(x,y)=0$.



                                  We have $nabla f=langle 1,0 rangle$ and $nabla g = langle 1-frac{1}{2sqrt{x}},1-frac{1}{2sqrt{y}} rangle$. At a maximum, we have $nabla g=lambda nabla f$ for some real number $lambda$. This implies that $1-frac{1}{2sqrt{y}}=0$, that is $y=frac{1}{4}$.



                                  Using the constraint, we have $x-sqrt{x}-frac{1}{4}=0$ which is equivalent to $a^2-a-frac{1}{4}$ and $sqrt{x}=ageqslant 0$. We obtain $a=frac{1+sqrt{2}}{2}$, so $x=frac{(1+sqrt{2})^2}{4}=frac{3+2sqrt{2}}{4}$.





                                  (sloppy) proof of the existence of the global maximum value: The constraint $x+y=sqrt{x}+sqrt{y}$ implies that $xgeqslant 0$ and $ygeqslant 0$. Moreover, if $x$ or $y$ is large enough (x>R or y>R for some $R$ that I am too lazy to determine explicitly :p), $x+y>sqrt{x}+sqrt{y}$, so we can assume that $x,y in [0,R]$.



                                  Now, the function $f(x,y)=x$ is continuous and the set $S=left{ (x,y)in [0,R]^2; middle|; g(x,y)=0; right}$ is compact (closed and bounded in ${mathbb R}^2$) so, by the Extreme Value Theorem, $f$ attains its global extreme values on $S$. The global minimum value is obviously obtained at $(0,0)$, while the other one is located by Lagrange's method.






                                  share|cite|improve this answer

























                                    up vote
                                    0
                                    down vote













                                    There are some caveats (we need to assume the existence of a maximum for example) but Lagrange's multipliers method work well here to find the value of the maximum:



                                    Let $f(x,y)=x$ and $g(x,y)=x-sqrt{x}+y-sqrt{y}$. We want to find the maximum value of $f$ under constraint $g(x,y)=0$.



                                    We have $nabla f=langle 1,0 rangle$ and $nabla g = langle 1-frac{1}{2sqrt{x}},1-frac{1}{2sqrt{y}} rangle$. At a maximum, we have $nabla g=lambda nabla f$ for some real number $lambda$. This implies that $1-frac{1}{2sqrt{y}}=0$, that is $y=frac{1}{4}$.



                                    Using the constraint, we have $x-sqrt{x}-frac{1}{4}=0$ which is equivalent to $a^2-a-frac{1}{4}$ and $sqrt{x}=ageqslant 0$. We obtain $a=frac{1+sqrt{2}}{2}$, so $x=frac{(1+sqrt{2})^2}{4}=frac{3+2sqrt{2}}{4}$.





                                    (sloppy) proof of the existence of the global maximum value: The constraint $x+y=sqrt{x}+sqrt{y}$ implies that $xgeqslant 0$ and $ygeqslant 0$. Moreover, if $x$ or $y$ is large enough (x>R or y>R for some $R$ that I am too lazy to determine explicitly :p), $x+y>sqrt{x}+sqrt{y}$, so we can assume that $x,y in [0,R]$.



                                    Now, the function $f(x,y)=x$ is continuous and the set $S=left{ (x,y)in [0,R]^2; middle|; g(x,y)=0; right}$ is compact (closed and bounded in ${mathbb R}^2$) so, by the Extreme Value Theorem, $f$ attains its global extreme values on $S$. The global minimum value is obviously obtained at $(0,0)$, while the other one is located by Lagrange's method.






                                    share|cite|improve this answer























                                      up vote
                                      0
                                      down vote










                                      up vote
                                      0
                                      down vote









                                      There are some caveats (we need to assume the existence of a maximum for example) but Lagrange's multipliers method work well here to find the value of the maximum:



                                      Let $f(x,y)=x$ and $g(x,y)=x-sqrt{x}+y-sqrt{y}$. We want to find the maximum value of $f$ under constraint $g(x,y)=0$.



                                      We have $nabla f=langle 1,0 rangle$ and $nabla g = langle 1-frac{1}{2sqrt{x}},1-frac{1}{2sqrt{y}} rangle$. At a maximum, we have $nabla g=lambda nabla f$ for some real number $lambda$. This implies that $1-frac{1}{2sqrt{y}}=0$, that is $y=frac{1}{4}$.



                                      Using the constraint, we have $x-sqrt{x}-frac{1}{4}=0$ which is equivalent to $a^2-a-frac{1}{4}$ and $sqrt{x}=ageqslant 0$. We obtain $a=frac{1+sqrt{2}}{2}$, so $x=frac{(1+sqrt{2})^2}{4}=frac{3+2sqrt{2}}{4}$.





                                      (sloppy) proof of the existence of the global maximum value: The constraint $x+y=sqrt{x}+sqrt{y}$ implies that $xgeqslant 0$ and $ygeqslant 0$. Moreover, if $x$ or $y$ is large enough (x>R or y>R for some $R$ that I am too lazy to determine explicitly :p), $x+y>sqrt{x}+sqrt{y}$, so we can assume that $x,y in [0,R]$.



                                      Now, the function $f(x,y)=x$ is continuous and the set $S=left{ (x,y)in [0,R]^2; middle|; g(x,y)=0; right}$ is compact (closed and bounded in ${mathbb R}^2$) so, by the Extreme Value Theorem, $f$ attains its global extreme values on $S$. The global minimum value is obviously obtained at $(0,0)$, while the other one is located by Lagrange's method.






                                      share|cite|improve this answer












                                      There are some caveats (we need to assume the existence of a maximum for example) but Lagrange's multipliers method work well here to find the value of the maximum:



                                      Let $f(x,y)=x$ and $g(x,y)=x-sqrt{x}+y-sqrt{y}$. We want to find the maximum value of $f$ under constraint $g(x,y)=0$.



                                      We have $nabla f=langle 1,0 rangle$ and $nabla g = langle 1-frac{1}{2sqrt{x}},1-frac{1}{2sqrt{y}} rangle$. At a maximum, we have $nabla g=lambda nabla f$ for some real number $lambda$. This implies that $1-frac{1}{2sqrt{y}}=0$, that is $y=frac{1}{4}$.



                                      Using the constraint, we have $x-sqrt{x}-frac{1}{4}=0$ which is equivalent to $a^2-a-frac{1}{4}$ and $sqrt{x}=ageqslant 0$. We obtain $a=frac{1+sqrt{2}}{2}$, so $x=frac{(1+sqrt{2})^2}{4}=frac{3+2sqrt{2}}{4}$.





                                      (sloppy) proof of the existence of the global maximum value: The constraint $x+y=sqrt{x}+sqrt{y}$ implies that $xgeqslant 0$ and $ygeqslant 0$. Moreover, if $x$ or $y$ is large enough (x>R or y>R for some $R$ that I am too lazy to determine explicitly :p), $x+y>sqrt{x}+sqrt{y}$, so we can assume that $x,y in [0,R]$.



                                      Now, the function $f(x,y)=x$ is continuous and the set $S=left{ (x,y)in [0,R]^2; middle|; g(x,y)=0; right}$ is compact (closed and bounded in ${mathbb R}^2$) so, by the Extreme Value Theorem, $f$ attains its global extreme values on $S$. The global minimum value is obviously obtained at $(0,0)$, while the other one is located by Lagrange's method.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 19 at 2:06









                                      Taladris

                                      4,63731832




                                      4,63731832






























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