Maximum value of $x$ when equality is given












4












$begingroup$


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










share|cite|improve this question











$endgroup$












  • $begingroup$
    "Find maximum value...of $x$"? What does this mean?
    $endgroup$
    – DonAntonio
    Nov 17 '18 at 20:17












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






  • 3




    $begingroup$
    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.
    $endgroup$
    – DonAntonio
    Nov 17 '18 at 20:20










  • $begingroup$
    The question makes no much sense, as @DonAntonio suggests.
    $endgroup$
    – Rebellos
    Nov 17 '18 at 20:22










  • $begingroup$
    Solve as quadric with respect to $sqrt{x}$ and then maximize the root value + the condition on integers. Would that not work?
    $endgroup$
    – Makina
    Nov 17 '18 at 20:23
















4












$begingroup$


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










share|cite|improve this question











$endgroup$












  • $begingroup$
    "Find maximum value...of $x$"? What does this mean?
    $endgroup$
    – DonAntonio
    Nov 17 '18 at 20:17












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






  • 3




    $begingroup$
    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.
    $endgroup$
    – DonAntonio
    Nov 17 '18 at 20:20










  • $begingroup$
    The question makes no much sense, as @DonAntonio suggests.
    $endgroup$
    – Rebellos
    Nov 17 '18 at 20:22










  • $begingroup$
    Solve as quadric with respect to $sqrt{x}$ and then maximize the root value + the condition on integers. Would that not work?
    $endgroup$
    – Makina
    Nov 17 '18 at 20:23














4












4








4





$begingroup$


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










share|cite|improve this question











$endgroup$




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







algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 18 '18 at 8:00









iBug

1449




1449










asked Nov 17 '18 at 20:13









mavericmaveric

69212




69212












  • $begingroup$
    "Find maximum value...of $x$"? What does this mean?
    $endgroup$
    – DonAntonio
    Nov 17 '18 at 20:17












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






  • 3




    $begingroup$
    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.
    $endgroup$
    – DonAntonio
    Nov 17 '18 at 20:20










  • $begingroup$
    The question makes no much sense, as @DonAntonio suggests.
    $endgroup$
    – Rebellos
    Nov 17 '18 at 20:22










  • $begingroup$
    Solve as quadric with respect to $sqrt{x}$ and then maximize the root value + the condition on integers. Would that not work?
    $endgroup$
    – Makina
    Nov 17 '18 at 20:23


















  • $begingroup$
    "Find maximum value...of $x$"? What does this mean?
    $endgroup$
    – DonAntonio
    Nov 17 '18 at 20:17












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






  • 3




    $begingroup$
    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.
    $endgroup$
    – DonAntonio
    Nov 17 '18 at 20:20










  • $begingroup$
    The question makes no much sense, as @DonAntonio suggests.
    $endgroup$
    – Rebellos
    Nov 17 '18 at 20:22










  • $begingroup$
    Solve as quadric with respect to $sqrt{x}$ and then maximize the root value + the condition on integers. Would that not work?
    $endgroup$
    – Makina
    Nov 17 '18 at 20:23
















$begingroup$
"Find maximum value...of $x$"? What does this mean?
$endgroup$
– DonAntonio
Nov 17 '18 at 20:17






$begingroup$
"Find maximum value...of $x$"? What does this mean?
$endgroup$
– DonAntonio
Nov 17 '18 at 20:17














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




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




3




3




$begingroup$
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.
$endgroup$
– DonAntonio
Nov 17 '18 at 20:20




$begingroup$
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.
$endgroup$
– DonAntonio
Nov 17 '18 at 20:20












$begingroup$
The question makes no much sense, as @DonAntonio suggests.
$endgroup$
– Rebellos
Nov 17 '18 at 20:22




$begingroup$
The question makes no much sense, as @DonAntonio suggests.
$endgroup$
– Rebellos
Nov 17 '18 at 20:22












$begingroup$
Solve as quadric with respect to $sqrt{x}$ and then maximize the root value + the condition on integers. Would that not work?
$endgroup$
– Makina
Nov 17 '18 at 20:23




$begingroup$
Solve as quadric with respect to $sqrt{x}$ and then maximize the root value + the condition on integers. Would that not work?
$endgroup$
– Makina
Nov 17 '18 at 20:23










7 Answers
7






active

oldest

votes


















4












$begingroup$

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









$endgroup$





















    3












    $begingroup$

    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









    $endgroup$





















      3












      $begingroup$

      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











      $endgroup$









      • 1




        $begingroup$
        dont understant what you trying to convey?
        $endgroup$
        – maveric
        Nov 17 '18 at 20:50



















      3












      $begingroup$

      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











      $endgroup$













      • $begingroup$
        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 $1.46$.
        $endgroup$
        – YiFan
        Nov 18 '18 at 6:25






      • 2




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






      • 1




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










      • $begingroup$
        +1. Nice method.
        $endgroup$
        – Taladris
        Nov 19 '18 at 0:45



















      1












      $begingroup$

      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









      $endgroup$













      • $begingroup$
        sorry the question is edited. it was real.
        $endgroup$
        – maveric
        Nov 17 '18 at 20:25










      • $begingroup$
        can we say then 1?
        $endgroup$
        – maveric
        Nov 17 '18 at 20:25










      • $begingroup$
        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...
        $endgroup$
        – DonAntonio
        Nov 17 '18 at 20:39



















      1












      $begingroup$

      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









      $endgroup$





















        0












        $begingroup$

        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









        $endgroup$













          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3002761%2fmaximum-value-of-x-when-equality-is-given%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          7 Answers
          7






          active

          oldest

          votes








          7 Answers
          7






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          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









          $endgroup$


















            4












            $begingroup$

            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









            $endgroup$
















              4












              4








              4





              $begingroup$

              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









              $endgroup$



              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 '18 at 20:25









              Mark BennetMark Bennet

              81k981179




              81k981179























                  3












                  $begingroup$

                  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









                  $endgroup$


















                    3












                    $begingroup$

                    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









                    $endgroup$
















                      3












                      3








                      3





                      $begingroup$

                      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









                      $endgroup$



                      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 '18 at 20:47









                      AndreiAndrei

                      11.9k21126




                      11.9k21126























                          3












                          $begingroup$

                          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











                          $endgroup$









                          • 1




                            $begingroup$
                            dont understant what you trying to convey?
                            $endgroup$
                            – maveric
                            Nov 17 '18 at 20:50
















                          3












                          $begingroup$

                          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











                          $endgroup$









                          • 1




                            $begingroup$
                            dont understant what you trying to convey?
                            $endgroup$
                            – maveric
                            Nov 17 '18 at 20:50














                          3












                          3








                          3





                          $begingroup$

                          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











                          $endgroup$



                          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 '18 at 6:21

























                          answered Nov 17 '18 at 20:47









                          zwimzwim

                          12k730




                          12k730








                          • 1




                            $begingroup$
                            dont understant what you trying to convey?
                            $endgroup$
                            – maveric
                            Nov 17 '18 at 20:50














                          • 1




                            $begingroup$
                            dont understant what you trying to convey?
                            $endgroup$
                            – maveric
                            Nov 17 '18 at 20:50








                          1




                          1




                          $begingroup$
                          dont understant what you trying to convey?
                          $endgroup$
                          – maveric
                          Nov 17 '18 at 20:50




                          $begingroup$
                          dont understant what you trying to convey?
                          $endgroup$
                          – maveric
                          Nov 17 '18 at 20:50











                          3












                          $begingroup$

                          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











                          $endgroup$













                          • $begingroup$
                            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 $1.46$.
                            $endgroup$
                            – YiFan
                            Nov 18 '18 at 6:25






                          • 2




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






                          • 1




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










                          • $begingroup$
                            +1. Nice method.
                            $endgroup$
                            – Taladris
                            Nov 19 '18 at 0:45
















                          3












                          $begingroup$

                          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











                          $endgroup$













                          • $begingroup$
                            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 $1.46$.
                            $endgroup$
                            – YiFan
                            Nov 18 '18 at 6:25






                          • 2




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






                          • 1




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










                          • $begingroup$
                            +1. Nice method.
                            $endgroup$
                            – Taladris
                            Nov 19 '18 at 0:45














                          3












                          3








                          3





                          $begingroup$

                          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











                          $endgroup$



                          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 '18 at 6:39

























                          answered Nov 18 '18 at 5:32









                          farruhotafarruhota

                          20.2k2738




                          20.2k2738












                          • $begingroup$
                            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 $1.46$.
                            $endgroup$
                            – YiFan
                            Nov 18 '18 at 6:25






                          • 2




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






                          • 1




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










                          • $begingroup$
                            +1. Nice method.
                            $endgroup$
                            – Taladris
                            Nov 19 '18 at 0:45


















                          • $begingroup$
                            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 $1.46$.
                            $endgroup$
                            – YiFan
                            Nov 18 '18 at 6:25






                          • 2




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






                          • 1




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










                          • $begingroup$
                            +1. Nice method.
                            $endgroup$
                            – Taladris
                            Nov 19 '18 at 0:45
















                          $begingroup$
                          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 $1.46$.
                          $endgroup$
                          – YiFan
                          Nov 18 '18 at 6:25




                          $begingroup$
                          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 $1.46$.
                          $endgroup$
                          – YiFan
                          Nov 18 '18 at 6:25




                          2




                          2




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




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




                          1




                          1




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




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












                          $begingroup$
                          +1. Nice method.
                          $endgroup$
                          – Taladris
                          Nov 19 '18 at 0:45




                          $begingroup$
                          +1. Nice method.
                          $endgroup$
                          – Taladris
                          Nov 19 '18 at 0:45











                          1












                          $begingroup$

                          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









                          $endgroup$













                          • $begingroup$
                            sorry the question is edited. it was real.
                            $endgroup$
                            – maveric
                            Nov 17 '18 at 20:25










                          • $begingroup$
                            can we say then 1?
                            $endgroup$
                            – maveric
                            Nov 17 '18 at 20:25










                          • $begingroup$
                            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...
                            $endgroup$
                            – DonAntonio
                            Nov 17 '18 at 20:39
















                          1












                          $begingroup$

                          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









                          $endgroup$













                          • $begingroup$
                            sorry the question is edited. it was real.
                            $endgroup$
                            – maveric
                            Nov 17 '18 at 20:25










                          • $begingroup$
                            can we say then 1?
                            $endgroup$
                            – maveric
                            Nov 17 '18 at 20:25










                          • $begingroup$
                            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...
                            $endgroup$
                            – DonAntonio
                            Nov 17 '18 at 20:39














                          1












                          1








                          1





                          $begingroup$

                          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









                          $endgroup$



                          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 '18 at 20:24









                          DonAntonioDonAntonio

                          178k1493229




                          178k1493229












                          • $begingroup$
                            sorry the question is edited. it was real.
                            $endgroup$
                            – maveric
                            Nov 17 '18 at 20:25










                          • $begingroup$
                            can we say then 1?
                            $endgroup$
                            – maveric
                            Nov 17 '18 at 20:25










                          • $begingroup$
                            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...
                            $endgroup$
                            – DonAntonio
                            Nov 17 '18 at 20:39


















                          • $begingroup$
                            sorry the question is edited. it was real.
                            $endgroup$
                            – maveric
                            Nov 17 '18 at 20:25










                          • $begingroup$
                            can we say then 1?
                            $endgroup$
                            – maveric
                            Nov 17 '18 at 20:25










                          • $begingroup$
                            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...
                            $endgroup$
                            – DonAntonio
                            Nov 17 '18 at 20:39
















                          $begingroup$
                          sorry the question is edited. it was real.
                          $endgroup$
                          – maveric
                          Nov 17 '18 at 20:25




                          $begingroup$
                          sorry the question is edited. it was real.
                          $endgroup$
                          – maveric
                          Nov 17 '18 at 20:25












                          $begingroup$
                          can we say then 1?
                          $endgroup$
                          – maveric
                          Nov 17 '18 at 20:25




                          $begingroup$
                          can we say then 1?
                          $endgroup$
                          – maveric
                          Nov 17 '18 at 20:25












                          $begingroup$
                          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...
                          $endgroup$
                          – DonAntonio
                          Nov 17 '18 at 20:39




                          $begingroup$
                          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...
                          $endgroup$
                          – DonAntonio
                          Nov 17 '18 at 20:39











                          1












                          $begingroup$

                          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









                          $endgroup$


















                            1












                            $begingroup$

                            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









                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              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









                              $endgroup$



                              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 '18 at 6:22









                              YiFanYiFan

                              3,2501424




                              3,2501424























                                  0












                                  $begingroup$

                                  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









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    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









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      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









                                      $endgroup$



                                      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 '18 at 2:06









                                      TaladrisTaladris

                                      4,69631933




                                      4,69631933






























                                          draft saved

                                          draft discarded




















































                                          Thanks for contributing an answer to Mathematics Stack Exchange!


                                          • Please be sure to answer the question. Provide details and share your research!

                                          But avoid



                                          • Asking for help, clarification, or responding to other answers.

                                          • Making statements based on opinion; back them up with references or personal experience.


                                          Use MathJax to format equations. MathJax reference.


                                          To learn more, see our tips on writing great answers.




                                          draft saved


                                          draft discarded














                                          StackExchange.ready(
                                          function () {
                                          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3002761%2fmaximum-value-of-x-when-equality-is-given%23new-answer', 'question_page');
                                          }
                                          );

                                          Post as a guest















                                          Required, but never shown





















































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown

































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown







                                          這個網誌中的熱門文章

                                          Xamarin.form Move up view when keyboard appear

                                          Post-Redirect-Get with Spring WebFlux and Thymeleaf

                                          Anylogic : not able to use stopDelay()