Maximizing $f$ in $mathbb{R}^3$












4












$begingroup$



Find the domain and the maximum value that the function
$$f(x,y,z)=frac{x+2y+3z}{sqrt{x^2+y^2+z^2}}$$
may attain in its domain.




I have found the domain of the function to be $mathbb{R^3backslashmathbf{0}}$. To maximize I differentiated in terms of $x,y,z$ having
$$f_x=frac{-2 x y-3 x z+y^2+z^2}{left(x^2+y^2+z^2right)^{3/2}},quad f_y=frac{2 x^2-x y+z (2 z-3 y)}{left(x^2+y^2+z^2right)^{3/2}},quad f_z=frac{3 left(x^2+y^2right)-z (x+2 y)}{left(x^2+y^2+z^2right)^{3/2}}$$
But to solve the system $f_x=0,f_y=0$ and $f_z=0$ is rather hard. What are the plausible values of $x,y,z$?






multivariable-calculus optimization partial-derivative maxima-minima cauchy-schwarz-inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    The easy way is to use CS here. However if you want to try to solve it the calculus way notice that the function is homogenous (you can scale the parameters with the same number and retain the same function value) so you are free to assume $x=1$ in the system $f_x=f_y=f_z = 0$. This leaves you with a simpler system to solve.
    $endgroup$
    – Winther
    Jan 8 at 0:46
















4












$begingroup$



Find the domain and the maximum value that the function
$$f(x,y,z)=frac{x+2y+3z}{sqrt{x^2+y^2+z^2}}$$
may attain in its domain.




I have found the domain of the function to be $mathbb{R^3backslashmathbf{0}}$. To maximize I differentiated in terms of $x,y,z$ having
$$f_x=frac{-2 x y-3 x z+y^2+z^2}{left(x^2+y^2+z^2right)^{3/2}},quad f_y=frac{2 x^2-x y+z (2 z-3 y)}{left(x^2+y^2+z^2right)^{3/2}},quad f_z=frac{3 left(x^2+y^2right)-z (x+2 y)}{left(x^2+y^2+z^2right)^{3/2}}$$
But to solve the system $f_x=0,f_y=0$ and $f_z=0$ is rather hard. What are the plausible values of $x,y,z$?






multivariable-calculus optimization partial-derivative maxima-minima cauchy-schwarz-inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    The easy way is to use CS here. However if you want to try to solve it the calculus way notice that the function is homogenous (you can scale the parameters with the same number and retain the same function value) so you are free to assume $x=1$ in the system $f_x=f_y=f_z = 0$. This leaves you with a simpler system to solve.
    $endgroup$
    – Winther
    Jan 8 at 0:46














4












4








4





$begingroup$



Find the domain and the maximum value that the function
$$f(x,y,z)=frac{x+2y+3z}{sqrt{x^2+y^2+z^2}}$$
may attain in its domain.




I have found the domain of the function to be $mathbb{R^3backslashmathbf{0}}$. To maximize I differentiated in terms of $x,y,z$ having
$$f_x=frac{-2 x y-3 x z+y^2+z^2}{left(x^2+y^2+z^2right)^{3/2}},quad f_y=frac{2 x^2-x y+z (2 z-3 y)}{left(x^2+y^2+z^2right)^{3/2}},quad f_z=frac{3 left(x^2+y^2right)-z (x+2 y)}{left(x^2+y^2+z^2right)^{3/2}}$$
But to solve the system $f_x=0,f_y=0$ and $f_z=0$ is rather hard. What are the plausible values of $x,y,z$?






multivariable-calculus optimization partial-derivative maxima-minima cauchy-schwarz-inequality







share|cite|improve this question











$endgroup$





Find the domain and the maximum value that the function
$$f(x,y,z)=frac{x+2y+3z}{sqrt{x^2+y^2+z^2}}$$
may attain in its domain.




I have found the domain of the function to be $mathbb{R^3backslashmathbf{0}}$. To maximize I differentiated in terms of $x,y,z$ having
$$f_x=frac{-2 x y-3 x z+y^2+z^2}{left(x^2+y^2+z^2right)^{3/2}},quad f_y=frac{2 x^2-x y+z (2 z-3 y)}{left(x^2+y^2+z^2right)^{3/2}},quad f_z=frac{3 left(x^2+y^2right)-z (x+2 y)}{left(x^2+y^2+z^2right)^{3/2}}$$
But to solve the system $f_x=0,f_y=0$ and $f_z=0$ is rather hard. What are the plausible values of $x,y,z$?






multivariable-calculus optimization partial-derivative maxima-minima cauchy-schwarz-inequality




multivariable-calculus optimization partial-derivative maxima-minima cauchy-schwarz-inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 10:29









Michael Rozenberg

100k1591192




100k1591192










asked Jan 8 at 0:30









DisPxyDisPxy

235




235












  • $begingroup$
    The easy way is to use CS here. However if you want to try to solve it the calculus way notice that the function is homogenous (you can scale the parameters with the same number and retain the same function value) so you are free to assume $x=1$ in the system $f_x=f_y=f_z = 0$. This leaves you with a simpler system to solve.
    $endgroup$
    – Winther
    Jan 8 at 0:46


















  • $begingroup$
    The easy way is to use CS here. However if you want to try to solve it the calculus way notice that the function is homogenous (you can scale the parameters with the same number and retain the same function value) so you are free to assume $x=1$ in the system $f_x=f_y=f_z = 0$. This leaves you with a simpler system to solve.
    $endgroup$
    – Winther
    Jan 8 at 0:46
















$begingroup$
The easy way is to use CS here. However if you want to try to solve it the calculus way notice that the function is homogenous (you can scale the parameters with the same number and retain the same function value) so you are free to assume $x=1$ in the system $f_x=f_y=f_z = 0$. This leaves you with a simpler system to solve.
$endgroup$
– Winther
Jan 8 at 0:46




$begingroup$
The easy way is to use CS here. However if you want to try to solve it the calculus way notice that the function is homogenous (you can scale the parameters with the same number and retain the same function value) so you are free to assume $x=1$ in the system $f_x=f_y=f_z = 0$. This leaves you with a simpler system to solve.
$endgroup$
– Winther
Jan 8 at 0:46










3 Answers
3






active

oldest

votes


















5












$begingroup$

Consider the vectors $,vec{u}=(x,y,z),$ and $,vec{v}=(1,2,3)$, then we can write
$$frac{x+2y+3z}{sqrt{x^2+y^2+z^2}}=frac{vec{u}boldsymbol{cdot}vec{v}}{Vertvec{u}Vert}=frac{Vertvec{u}VertVertvec{v}Vertcos(alpha)}{Vertvec{u}Vert}=Vertvec{v}Vertcos(alpha)=sqrt{1^2+2^2+3^2}cos(alpha
)=sqrt{14}cos(alpha)$$



When is the last expression maximized? When $alpha=kpi,,kinmathbb{Z}$, thus the maximum value of $f$ is $sqrt{14}$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    I think it should be $2kpi$
    $endgroup$
    – DisPxy
    Jan 8 at 10:07



















4












$begingroup$

By C-S $$(x^2+y^2+z^2)(1^2+2^2+3^2)geq(x+2y+3z)^2,$$
which gives
$$-sqrt{14}leqfrac{x+2y+3z}{sqrt{x^2+y^2+z^2}}leqsqrt{14}.$$
The equality occurs for $(x,y,z)||(1,2,3),$ which gives that we got a maximal value and the minimal value.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What does C-S mean?
    $endgroup$
    – DisPxy
    Jan 8 at 0:46






  • 1




    $begingroup$
    It's Cauchy-Schwarz inequality.
    $endgroup$
    – Michael Rozenberg
    Jan 8 at 0:47



















2












$begingroup$

Consider the line
$x = t, y = 2t, z = 3t$



Along this line.



$f(t,2t,3t) = frac {14 t}{sqrt{14 t^2}} = sqrt {14}$



let's find some orthogonal vectors.



$mathbf u = (1,2,3)\
mathbf v = (2,-1,0)\
mathbf w = (0,3,-2)$



Any point in $mathbb R^3$ is some linear combination



$c_1mathbf u + c_2mathbf v+ c_3mathbf w$



$f(c_1mathbf u + c_2mathbf v+ c_3mathbf w) = frac {14c_1}{sqrt{14c_1^2 + 5c_2^2 + 13c_3^2}}\
|f|le sqrt 14 text{ sgn}(c_1)$






share|cite|improve this answer









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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    Consider the vectors $,vec{u}=(x,y,z),$ and $,vec{v}=(1,2,3)$, then we can write
    $$frac{x+2y+3z}{sqrt{x^2+y^2+z^2}}=frac{vec{u}boldsymbol{cdot}vec{v}}{Vertvec{u}Vert}=frac{Vertvec{u}VertVertvec{v}Vertcos(alpha)}{Vertvec{u}Vert}=Vertvec{v}Vertcos(alpha)=sqrt{1^2+2^2+3^2}cos(alpha
    )=sqrt{14}cos(alpha)$$



    When is the last expression maximized? When $alpha=kpi,,kinmathbb{Z}$, thus the maximum value of $f$ is $sqrt{14}$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      I think it should be $2kpi$
      $endgroup$
      – DisPxy
      Jan 8 at 10:07
















    5












    $begingroup$

    Consider the vectors $,vec{u}=(x,y,z),$ and $,vec{v}=(1,2,3)$, then we can write
    $$frac{x+2y+3z}{sqrt{x^2+y^2+z^2}}=frac{vec{u}boldsymbol{cdot}vec{v}}{Vertvec{u}Vert}=frac{Vertvec{u}VertVertvec{v}Vertcos(alpha)}{Vertvec{u}Vert}=Vertvec{v}Vertcos(alpha)=sqrt{1^2+2^2+3^2}cos(alpha
    )=sqrt{14}cos(alpha)$$



    When is the last expression maximized? When $alpha=kpi,,kinmathbb{Z}$, thus the maximum value of $f$ is $sqrt{14}$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      I think it should be $2kpi$
      $endgroup$
      – DisPxy
      Jan 8 at 10:07














    5












    5








    5





    $begingroup$

    Consider the vectors $,vec{u}=(x,y,z),$ and $,vec{v}=(1,2,3)$, then we can write
    $$frac{x+2y+3z}{sqrt{x^2+y^2+z^2}}=frac{vec{u}boldsymbol{cdot}vec{v}}{Vertvec{u}Vert}=frac{Vertvec{u}VertVertvec{v}Vertcos(alpha)}{Vertvec{u}Vert}=Vertvec{v}Vertcos(alpha)=sqrt{1^2+2^2+3^2}cos(alpha
    )=sqrt{14}cos(alpha)$$



    When is the last expression maximized? When $alpha=kpi,,kinmathbb{Z}$, thus the maximum value of $f$ is $sqrt{14}$.






    share|cite|improve this answer









    $endgroup$



    Consider the vectors $,vec{u}=(x,y,z),$ and $,vec{v}=(1,2,3)$, then we can write
    $$frac{x+2y+3z}{sqrt{x^2+y^2+z^2}}=frac{vec{u}boldsymbol{cdot}vec{v}}{Vertvec{u}Vert}=frac{Vertvec{u}VertVertvec{v}Vertcos(alpha)}{Vertvec{u}Vert}=Vertvec{v}Vertcos(alpha)=sqrt{1^2+2^2+3^2}cos(alpha
    )=sqrt{14}cos(alpha)$$



    When is the last expression maximized? When $alpha=kpi,,kinmathbb{Z}$, thus the maximum value of $f$ is $sqrt{14}$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 8 at 0:39









    Denis28Denis28

    884218




    884218








    • 1




      $begingroup$
      I think it should be $2kpi$
      $endgroup$
      – DisPxy
      Jan 8 at 10:07














    • 1




      $begingroup$
      I think it should be $2kpi$
      $endgroup$
      – DisPxy
      Jan 8 at 10:07








    1




    1




    $begingroup$
    I think it should be $2kpi$
    $endgroup$
    – DisPxy
    Jan 8 at 10:07




    $begingroup$
    I think it should be $2kpi$
    $endgroup$
    – DisPxy
    Jan 8 at 10:07











    4












    $begingroup$

    By C-S $$(x^2+y^2+z^2)(1^2+2^2+3^2)geq(x+2y+3z)^2,$$
    which gives
    $$-sqrt{14}leqfrac{x+2y+3z}{sqrt{x^2+y^2+z^2}}leqsqrt{14}.$$
    The equality occurs for $(x,y,z)||(1,2,3),$ which gives that we got a maximal value and the minimal value.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      What does C-S mean?
      $endgroup$
      – DisPxy
      Jan 8 at 0:46






    • 1




      $begingroup$
      It's Cauchy-Schwarz inequality.
      $endgroup$
      – Michael Rozenberg
      Jan 8 at 0:47
















    4












    $begingroup$

    By C-S $$(x^2+y^2+z^2)(1^2+2^2+3^2)geq(x+2y+3z)^2,$$
    which gives
    $$-sqrt{14}leqfrac{x+2y+3z}{sqrt{x^2+y^2+z^2}}leqsqrt{14}.$$
    The equality occurs for $(x,y,z)||(1,2,3),$ which gives that we got a maximal value and the minimal value.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      What does C-S mean?
      $endgroup$
      – DisPxy
      Jan 8 at 0:46






    • 1




      $begingroup$
      It's Cauchy-Schwarz inequality.
      $endgroup$
      – Michael Rozenberg
      Jan 8 at 0:47














    4












    4








    4





    $begingroup$

    By C-S $$(x^2+y^2+z^2)(1^2+2^2+3^2)geq(x+2y+3z)^2,$$
    which gives
    $$-sqrt{14}leqfrac{x+2y+3z}{sqrt{x^2+y^2+z^2}}leqsqrt{14}.$$
    The equality occurs for $(x,y,z)||(1,2,3),$ which gives that we got a maximal value and the minimal value.






    share|cite|improve this answer









    $endgroup$



    By C-S $$(x^2+y^2+z^2)(1^2+2^2+3^2)geq(x+2y+3z)^2,$$
    which gives
    $$-sqrt{14}leqfrac{x+2y+3z}{sqrt{x^2+y^2+z^2}}leqsqrt{14}.$$
    The equality occurs for $(x,y,z)||(1,2,3),$ which gives that we got a maximal value and the minimal value.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 8 at 0:35









    Michael RozenbergMichael Rozenberg

    100k1591192




    100k1591192












    • $begingroup$
      What does C-S mean?
      $endgroup$
      – DisPxy
      Jan 8 at 0:46






    • 1




      $begingroup$
      It's Cauchy-Schwarz inequality.
      $endgroup$
      – Michael Rozenberg
      Jan 8 at 0:47


















    • $begingroup$
      What does C-S mean?
      $endgroup$
      – DisPxy
      Jan 8 at 0:46






    • 1




      $begingroup$
      It's Cauchy-Schwarz inequality.
      $endgroup$
      – Michael Rozenberg
      Jan 8 at 0:47
















    $begingroup$
    What does C-S mean?
    $endgroup$
    – DisPxy
    Jan 8 at 0:46




    $begingroup$
    What does C-S mean?
    $endgroup$
    – DisPxy
    Jan 8 at 0:46




    1




    1




    $begingroup$
    It's Cauchy-Schwarz inequality.
    $endgroup$
    – Michael Rozenberg
    Jan 8 at 0:47




    $begingroup$
    It's Cauchy-Schwarz inequality.
    $endgroup$
    – Michael Rozenberg
    Jan 8 at 0:47











    2












    $begingroup$

    Consider the line
    $x = t, y = 2t, z = 3t$



    Along this line.



    $f(t,2t,3t) = frac {14 t}{sqrt{14 t^2}} = sqrt {14}$



    let's find some orthogonal vectors.



    $mathbf u = (1,2,3)\
    mathbf v = (2,-1,0)\
    mathbf w = (0,3,-2)$



    Any point in $mathbb R^3$ is some linear combination



    $c_1mathbf u + c_2mathbf v+ c_3mathbf w$



    $f(c_1mathbf u + c_2mathbf v+ c_3mathbf w) = frac {14c_1}{sqrt{14c_1^2 + 5c_2^2 + 13c_3^2}}\
    |f|le sqrt 14 text{ sgn}(c_1)$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Consider the line
      $x = t, y = 2t, z = 3t$



      Along this line.



      $f(t,2t,3t) = frac {14 t}{sqrt{14 t^2}} = sqrt {14}$



      let's find some orthogonal vectors.



      $mathbf u = (1,2,3)\
      mathbf v = (2,-1,0)\
      mathbf w = (0,3,-2)$



      Any point in $mathbb R^3$ is some linear combination



      $c_1mathbf u + c_2mathbf v+ c_3mathbf w$



      $f(c_1mathbf u + c_2mathbf v+ c_3mathbf w) = frac {14c_1}{sqrt{14c_1^2 + 5c_2^2 + 13c_3^2}}\
      |f|le sqrt 14 text{ sgn}(c_1)$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Consider the line
        $x = t, y = 2t, z = 3t$



        Along this line.



        $f(t,2t,3t) = frac {14 t}{sqrt{14 t^2}} = sqrt {14}$



        let's find some orthogonal vectors.



        $mathbf u = (1,2,3)\
        mathbf v = (2,-1,0)\
        mathbf w = (0,3,-2)$



        Any point in $mathbb R^3$ is some linear combination



        $c_1mathbf u + c_2mathbf v+ c_3mathbf w$



        $f(c_1mathbf u + c_2mathbf v+ c_3mathbf w) = frac {14c_1}{sqrt{14c_1^2 + 5c_2^2 + 13c_3^2}}\
        |f|le sqrt 14 text{ sgn}(c_1)$






        share|cite|improve this answer









        $endgroup$



        Consider the line
        $x = t, y = 2t, z = 3t$



        Along this line.



        $f(t,2t,3t) = frac {14 t}{sqrt{14 t^2}} = sqrt {14}$



        let's find some orthogonal vectors.



        $mathbf u = (1,2,3)\
        mathbf v = (2,-1,0)\
        mathbf w = (0,3,-2)$



        Any point in $mathbb R^3$ is some linear combination



        $c_1mathbf u + c_2mathbf v+ c_3mathbf w$



        $f(c_1mathbf u + c_2mathbf v+ c_3mathbf w) = frac {14c_1}{sqrt{14c_1^2 + 5c_2^2 + 13c_3^2}}\
        |f|le sqrt 14 text{ sgn}(c_1)$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 1:57









        Doug MDoug M

        44.6k31854




        44.6k31854






























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            in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

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            1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
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