Use Lagrange multiplier to find extrema of $f(x,y,z,t)$ subject to the stated constraints.












0















Use Lagrange multiplier to find extrema of $f$ subject to the stated constraints.



$f(x,y,z,t)=xyzt$



$x-z=2$ and $y^2+t=4$




My attempt:



Let $g(x,z)=x-z-2=0$ and $h(y,t)=y^2+t-4=0$



$nabla f=lambda nabla g+mu nabla h$



$(yzt)i+(xzt)j+(xyt)k+(xyz)l=lambda (i-k)+mu (2y j + l)$



Therefore,



$xyzt=lambda x$



$xyzt=2mu y^2$



$xyzt=-lambda z$



$xyzt=mu t$



$x-z-2=0$



$y^2+t-4=0$



Now, we wnat to solve this system, and by add the first four equations, we have



$frac{7}{2} xyzt=lambda (x-z)+mu (y^2+t)$



$=2 lambda +4mu=2(lambda +2mu)$



$xyzt=frac{4}{7} (lambda+mu)$



Then,



$lambda x= frac{4}{7} (lambda+mu)$



$mu y^2= frac{2}{7} (lambda+mu)$



$lambda z= frac{-4}{7} (lambda+mu)$



$mu t=frac{4}{7} (lambda+mu)$



Now what can I do? Can I divide these equations by $lambda$ or $mu$ to find $x,y,z$ and $t$? But I don’t know what the value of $lambda$ and $mu$ it may be will zero!



Please help :) thanks.






calculus lagrange-multiplier







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    0















    Use Lagrange multiplier to find extrema of $f$ subject to the stated constraints.



    $f(x,y,z,t)=xyzt$



    $x-z=2$ and $y^2+t=4$




    My attempt:



    Let $g(x,z)=x-z-2=0$ and $h(y,t)=y^2+t-4=0$



    $nabla f=lambda nabla g+mu nabla h$



    $(yzt)i+(xzt)j+(xyt)k+(xyz)l=lambda (i-k)+mu (2y j + l)$



    Therefore,



    $xyzt=lambda x$



    $xyzt=2mu y^2$



    $xyzt=-lambda z$



    $xyzt=mu t$



    $x-z-2=0$



    $y^2+t-4=0$



    Now, we wnat to solve this system, and by add the first four equations, we have



    $frac{7}{2} xyzt=lambda (x-z)+mu (y^2+t)$



    $=2 lambda +4mu=2(lambda +2mu)$



    $xyzt=frac{4}{7} (lambda+mu)$



    Then,



    $lambda x= frac{4}{7} (lambda+mu)$



    $mu y^2= frac{2}{7} (lambda+mu)$



    $lambda z= frac{-4}{7} (lambda+mu)$



    $mu t=frac{4}{7} (lambda+mu)$



    Now what can I do? Can I divide these equations by $lambda$ or $mu$ to find $x,y,z$ and $t$? But I don’t know what the value of $lambda$ and $mu$ it may be will zero!



    Please help :) thanks.






    calculus lagrange-multiplier







    share|cite|improve this question

























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      Use Lagrange multiplier to find extrema of $f$ subject to the stated constraints.



      $f(x,y,z,t)=xyzt$



      $x-z=2$ and $y^2+t=4$




      My attempt:



      Let $g(x,z)=x-z-2=0$ and $h(y,t)=y^2+t-4=0$



      $nabla f=lambda nabla g+mu nabla h$



      $(yzt)i+(xzt)j+(xyt)k+(xyz)l=lambda (i-k)+mu (2y j + l)$



      Therefore,



      $xyzt=lambda x$



      $xyzt=2mu y^2$



      $xyzt=-lambda z$



      $xyzt=mu t$



      $x-z-2=0$



      $y^2+t-4=0$



      Now, we wnat to solve this system, and by add the first four equations, we have



      $frac{7}{2} xyzt=lambda (x-z)+mu (y^2+t)$



      $=2 lambda +4mu=2(lambda +2mu)$



      $xyzt=frac{4}{7} (lambda+mu)$



      Then,



      $lambda x= frac{4}{7} (lambda+mu)$



      $mu y^2= frac{2}{7} (lambda+mu)$



      $lambda z= frac{-4}{7} (lambda+mu)$



      $mu t=frac{4}{7} (lambda+mu)$



      Now what can I do? Can I divide these equations by $lambda$ or $mu$ to find $x,y,z$ and $t$? But I don’t know what the value of $lambda$ and $mu$ it may be will zero!



      Please help :) thanks.






      calculus lagrange-multiplier







      share|cite|improve this question














      Use Lagrange multiplier to find extrema of $f$ subject to the stated constraints.



      $f(x,y,z,t)=xyzt$



      $x-z=2$ and $y^2+t=4$




      My attempt:



      Let $g(x,z)=x-z-2=0$ and $h(y,t)=y^2+t-4=0$



      $nabla f=lambda nabla g+mu nabla h$



      $(yzt)i+(xzt)j+(xyt)k+(xyz)l=lambda (i-k)+mu (2y j + l)$



      Therefore,



      $xyzt=lambda x$



      $xyzt=2mu y^2$



      $xyzt=-lambda z$



      $xyzt=mu t$



      $x-z-2=0$



      $y^2+t-4=0$



      Now, we wnat to solve this system, and by add the first four equations, we have



      $frac{7}{2} xyzt=lambda (x-z)+mu (y^2+t)$



      $=2 lambda +4mu=2(lambda +2mu)$



      $xyzt=frac{4}{7} (lambda+mu)$



      Then,



      $lambda x= frac{4}{7} (lambda+mu)$



      $mu y^2= frac{2}{7} (lambda+mu)$



      $lambda z= frac{-4}{7} (lambda+mu)$



      $mu t=frac{4}{7} (lambda+mu)$



      Now what can I do? Can I divide these equations by $lambda$ or $mu$ to find $x,y,z$ and $t$? But I don’t know what the value of $lambda$ and $mu$ it may be will zero!



      Please help :) thanks.






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      calculus lagrange-multiplier






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      asked Dec 1 at 12:17









      Dima

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          The stationary points can be found as follows:



          We have the following first order conditions:



          $$yzt = lambda tag{1}$$



          $$xzt=2mu ytag{2}$$



          $$xyt=-lambdatag{3}$$



          $$xyz=mu tag{4}$$



          $$x-z=2tag{5}$$



          $$y^2+t=4 tag{6}$$



          From $(1)$ and $(3)$,



          $$yzt=-xyt iff (x+z)yt=0$$
          From $(2)$ and $(4)$,
          $$xzt=2xy^2z iff xz(t-2y^2)=0$$



          We consider the case where the objective function is non-zero.



          $$x+z=0$$ and $$x-z=2$$



          Then we have $(x,z)=(1,-1)$.



          Also if $$t-2y^2=0$$



          and $$y^2+t=4$$



          Then we have $$2y^2=4-y^2 iff y^2=frac{4}{3}$$



          $$(t,y)=left( frac83, pmfrac{2}{sqrt3}right)$$



          Remark: We still have to examine if the function can be unbounded.



          If we let $x=2, z=1$, $t=4-y^2$, then the objective function becomes $2y(4-y^2)$, it is indeed unbounded.






          share|cite|improve this answer























          • Thank you so so so much prof.
            – Dima
            Dec 1 at 12:46






          • 1




            Note that I haven't consider some stationary points, those are the ones that make the objective function $0$.
            – Siong Thye Goh
            Dec 1 at 12:48











          Your Answer





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          1 Answer
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          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          2














          The stationary points can be found as follows:



          We have the following first order conditions:



          $$yzt = lambda tag{1}$$



          $$xzt=2mu ytag{2}$$



          $$xyt=-lambdatag{3}$$



          $$xyz=mu tag{4}$$



          $$x-z=2tag{5}$$



          $$y^2+t=4 tag{6}$$



          From $(1)$ and $(3)$,



          $$yzt=-xyt iff (x+z)yt=0$$
          From $(2)$ and $(4)$,
          $$xzt=2xy^2z iff xz(t-2y^2)=0$$



          We consider the case where the objective function is non-zero.



          $$x+z=0$$ and $$x-z=2$$



          Then we have $(x,z)=(1,-1)$.



          Also if $$t-2y^2=0$$



          and $$y^2+t=4$$



          Then we have $$2y^2=4-y^2 iff y^2=frac{4}{3}$$



          $$(t,y)=left( frac83, pmfrac{2}{sqrt3}right)$$



          Remark: We still have to examine if the function can be unbounded.



          If we let $x=2, z=1$, $t=4-y^2$, then the objective function becomes $2y(4-y^2)$, it is indeed unbounded.






          share|cite|improve this answer























          • Thank you so so so much prof.
            – Dima
            Dec 1 at 12:46






          • 1




            Note that I haven't consider some stationary points, those are the ones that make the objective function $0$.
            – Siong Thye Goh
            Dec 1 at 12:48
















          2














          The stationary points can be found as follows:



          We have the following first order conditions:



          $$yzt = lambda tag{1}$$



          $$xzt=2mu ytag{2}$$



          $$xyt=-lambdatag{3}$$



          $$xyz=mu tag{4}$$



          $$x-z=2tag{5}$$



          $$y^2+t=4 tag{6}$$



          From $(1)$ and $(3)$,



          $$yzt=-xyt iff (x+z)yt=0$$
          From $(2)$ and $(4)$,
          $$xzt=2xy^2z iff xz(t-2y^2)=0$$



          We consider the case where the objective function is non-zero.



          $$x+z=0$$ and $$x-z=2$$



          Then we have $(x,z)=(1,-1)$.



          Also if $$t-2y^2=0$$



          and $$y^2+t=4$$



          Then we have $$2y^2=4-y^2 iff y^2=frac{4}{3}$$



          $$(t,y)=left( frac83, pmfrac{2}{sqrt3}right)$$



          Remark: We still have to examine if the function can be unbounded.



          If we let $x=2, z=1$, $t=4-y^2$, then the objective function becomes $2y(4-y^2)$, it is indeed unbounded.






          share|cite|improve this answer























          • Thank you so so so much prof.
            – Dima
            Dec 1 at 12:46






          • 1




            Note that I haven't consider some stationary points, those are the ones that make the objective function $0$.
            – Siong Thye Goh
            Dec 1 at 12:48














          2












          2








          2






          The stationary points can be found as follows:



          We have the following first order conditions:



          $$yzt = lambda tag{1}$$



          $$xzt=2mu ytag{2}$$



          $$xyt=-lambdatag{3}$$



          $$xyz=mu tag{4}$$



          $$x-z=2tag{5}$$



          $$y^2+t=4 tag{6}$$



          From $(1)$ and $(3)$,



          $$yzt=-xyt iff (x+z)yt=0$$
          From $(2)$ and $(4)$,
          $$xzt=2xy^2z iff xz(t-2y^2)=0$$



          We consider the case where the objective function is non-zero.



          $$x+z=0$$ and $$x-z=2$$



          Then we have $(x,z)=(1,-1)$.



          Also if $$t-2y^2=0$$



          and $$y^2+t=4$$



          Then we have $$2y^2=4-y^2 iff y^2=frac{4}{3}$$



          $$(t,y)=left( frac83, pmfrac{2}{sqrt3}right)$$



          Remark: We still have to examine if the function can be unbounded.



          If we let $x=2, z=1$, $t=4-y^2$, then the objective function becomes $2y(4-y^2)$, it is indeed unbounded.






          share|cite|improve this answer














          The stationary points can be found as follows:



          We have the following first order conditions:



          $$yzt = lambda tag{1}$$



          $$xzt=2mu ytag{2}$$



          $$xyt=-lambdatag{3}$$



          $$xyz=mu tag{4}$$



          $$x-z=2tag{5}$$



          $$y^2+t=4 tag{6}$$



          From $(1)$ and $(3)$,



          $$yzt=-xyt iff (x+z)yt=0$$
          From $(2)$ and $(4)$,
          $$xzt=2xy^2z iff xz(t-2y^2)=0$$



          We consider the case where the objective function is non-zero.



          $$x+z=0$$ and $$x-z=2$$



          Then we have $(x,z)=(1,-1)$.



          Also if $$t-2y^2=0$$



          and $$y^2+t=4$$



          Then we have $$2y^2=4-y^2 iff y^2=frac{4}{3}$$



          $$(t,y)=left( frac83, pmfrac{2}{sqrt3}right)$$



          Remark: We still have to examine if the function can be unbounded.



          If we let $x=2, z=1$, $t=4-y^2$, then the objective function becomes $2y(4-y^2)$, it is indeed unbounded.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 1 at 12:41

























          answered Dec 1 at 12:33









          Siong Thye Goh

          98.9k1464116




          98.9k1464116












          • Thank you so so so much prof.
            – Dima
            Dec 1 at 12:46






          • 1




            Note that I haven't consider some stationary points, those are the ones that make the objective function $0$.
            – Siong Thye Goh
            Dec 1 at 12:48


















          • Thank you so so so much prof.
            – Dima
            Dec 1 at 12:46






          • 1




            Note that I haven't consider some stationary points, those are the ones that make the objective function $0$.
            – Siong Thye Goh
            Dec 1 at 12:48
















          Thank you so so so much prof.
          – Dima
          Dec 1 at 12:46




          Thank you so so so much prof.
          – Dima
          Dec 1 at 12:46




          1




          1




          Note that I haven't consider some stationary points, those are the ones that make the objective function $0$.
          – Siong Thye Goh
          Dec 1 at 12:48




          Note that I haven't consider some stationary points, those are the ones that make the objective function $0$.
          – Siong Thye Goh
          Dec 1 at 12:48


















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