Use Lagrange multipliers to find the max and min of the function $f(x,y)=xe^y$ subject to the constraint...












1












$begingroup$



Use Lagrange multipliers to find the max and min of the function $f(x,y)=xe^y$ subject to the constraint $x^2+y^2=6$.




My attempt:



I tried to find both partials and set them equal to $lambda$ times the partial of the constraint and got the following equations



$$e^y=2x(lambda)$$
$$x(e^y)=2y(lambda) $$



I then solved for $x$ and $y$ and got :
$$x=frac{e^y}{2lambda}$$
$$y=frac{xe^y}{2lambda}$$



I then plugged those into the original equation and got



$$frac{(x^2+1)(e^(2y) )}{4lambda^2}=6$$



I'm confused where to go from here and any help will be greatly appreciated






multivariable-calculus lagrange-multiplier







share|cite|improve this question











$endgroup$












  • $begingroup$
    Try to eliminate $lambda$ from the two equations.
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 23:08










  • $begingroup$
    How would I go about doing that?
    $endgroup$
    – Emily
    Dec 2 '18 at 23:10
















1












$begingroup$



Use Lagrange multipliers to find the max and min of the function $f(x,y)=xe^y$ subject to the constraint $x^2+y^2=6$.




My attempt:



I tried to find both partials and set them equal to $lambda$ times the partial of the constraint and got the following equations



$$e^y=2x(lambda)$$
$$x(e^y)=2y(lambda) $$



I then solved for $x$ and $y$ and got :
$$x=frac{e^y}{2lambda}$$
$$y=frac{xe^y}{2lambda}$$



I then plugged those into the original equation and got



$$frac{(x^2+1)(e^(2y) )}{4lambda^2}=6$$



I'm confused where to go from here and any help will be greatly appreciated






multivariable-calculus lagrange-multiplier







share|cite|improve this question











$endgroup$












  • $begingroup$
    Try to eliminate $lambda$ from the two equations.
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 23:08










  • $begingroup$
    How would I go about doing that?
    $endgroup$
    – Emily
    Dec 2 '18 at 23:10














1












1








1


1



$begingroup$



Use Lagrange multipliers to find the max and min of the function $f(x,y)=xe^y$ subject to the constraint $x^2+y^2=6$.




My attempt:



I tried to find both partials and set them equal to $lambda$ times the partial of the constraint and got the following equations



$$e^y=2x(lambda)$$
$$x(e^y)=2y(lambda) $$



I then solved for $x$ and $y$ and got :
$$x=frac{e^y}{2lambda}$$
$$y=frac{xe^y}{2lambda}$$



I then plugged those into the original equation and got



$$frac{(x^2+1)(e^(2y) )}{4lambda^2}=6$$



I'm confused where to go from here and any help will be greatly appreciated






multivariable-calculus lagrange-multiplier







share|cite|improve this question











$endgroup$





Use Lagrange multipliers to find the max and min of the function $f(x,y)=xe^y$ subject to the constraint $x^2+y^2=6$.




My attempt:



I tried to find both partials and set them equal to $lambda$ times the partial of the constraint and got the following equations



$$e^y=2x(lambda)$$
$$x(e^y)=2y(lambda) $$



I then solved for $x$ and $y$ and got :
$$x=frac{e^y}{2lambda}$$
$$y=frac{xe^y}{2lambda}$$



I then plugged those into the original equation and got



$$frac{(x^2+1)(e^(2y) )}{4lambda^2}=6$$



I'm confused where to go from here and any help will be greatly appreciated






multivariable-calculus lagrange-multiplier




multivariable-calculus lagrange-multiplier






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 23:02







Emily

















asked Dec 2 '18 at 22:56









EmilyEmily

133




133












  • $begingroup$
    Try to eliminate $lambda$ from the two equations.
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 23:08










  • $begingroup$
    How would I go about doing that?
    $endgroup$
    – Emily
    Dec 2 '18 at 23:10


















  • $begingroup$
    Try to eliminate $lambda$ from the two equations.
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 23:08










  • $begingroup$
    How would I go about doing that?
    $endgroup$
    – Emily
    Dec 2 '18 at 23:10
















$begingroup$
Try to eliminate $lambda$ from the two equations.
$endgroup$
– Tito Eliatron
Dec 2 '18 at 23:08




$begingroup$
Try to eliminate $lambda$ from the two equations.
$endgroup$
– Tito Eliatron
Dec 2 '18 at 23:08












$begingroup$
How would I go about doing that?
$endgroup$
– Emily
Dec 2 '18 at 23:10




$begingroup$
How would I go about doing that?
$endgroup$
– Emily
Dec 2 '18 at 23:10










3 Answers
3






active

oldest

votes


















0












$begingroup$

We have,
$$e^y = 2 lambda x$$
$$x e^y = 2 lambda y$$
thus,
$$x = frac{2 lambda y}{e^y} = frac{2 lambda y}{2 lambda x} = frac{y}{x}$$
finally,
$$x^2 = y$$
Now we replace into the original (constraint) equation
$$y^2 + y - 6 = 0$$
You have now the constraint in the form of a quadratic equation, you can easily find the value of $y$ through solving the equation.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much I just found the roots to be 2 and -3 but I'm confused what to do with those?
    $endgroup$
    – Emily
    Dec 2 '18 at 23:43










  • $begingroup$
    You repalce the values of $y$ that you have found in the constraint equation & you end up getting two points $(x_1,y_1)$ & $(x_2,y_2)$, one of these two points is the local max. while the other is the local min., to verify which one is the local max. & which one is the local min. you calculate the second order partial derivatives of $f(x,y)$ because it's a multivariate function you will calculate the Hessian matrix & its determinant to find out which of the two points is local min. & which is the local max.,more here: analyzemath.com/calculus/multivariable/maxima_minima.html
    $endgroup$
    – Blg Khalil
    Dec 3 '18 at 2:28





















0












$begingroup$

Following your calculations:
$$2lambda=e^y/x$$
$$2lambda=xe^y/y$$
so
$$e^y/x=xe^y/y iff ye^y=x^2e^y iff y=x^2$$ where in the last equivalency we use that $e^yne0$.



Now, use this information with your constraint to get a bicuadratic equation on $x$ that you can solve alone.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Without Lagrange Multipliers



    Calling



    $$
    x = rcos(theta)\
    y = rsin(theta)
    $$



    we have the equivalent problem



    $$
    maxmin_{theta}f(theta) = sqrt{6}cos(theta) e^{sqrt 6sin(theta)}
    $$



    and



    $$
    f'(theta) = sqrt{6}e^{sqrt{6} sin (theta )} left(sqrt{6} cos ^2(theta )-sin (theta )right)to sqrt{6} cos ^2(theta )-sin (theta ) = 0
    $$



    etc.



    NOTE



    This result is equivalent to the system



    $$
    x^2+y^2=6\
    x^2-y = 0
    $$






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      We have,
      $$e^y = 2 lambda x$$
      $$x e^y = 2 lambda y$$
      thus,
      $$x = frac{2 lambda y}{e^y} = frac{2 lambda y}{2 lambda x} = frac{y}{x}$$
      finally,
      $$x^2 = y$$
      Now we replace into the original (constraint) equation
      $$y^2 + y - 6 = 0$$
      You have now the constraint in the form of a quadratic equation, you can easily find the value of $y$ through solving the equation.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thank you so much I just found the roots to be 2 and -3 but I'm confused what to do with those?
        $endgroup$
        – Emily
        Dec 2 '18 at 23:43










      • $begingroup$
        You repalce the values of $y$ that you have found in the constraint equation & you end up getting two points $(x_1,y_1)$ & $(x_2,y_2)$, one of these two points is the local max. while the other is the local min., to verify which one is the local max. & which one is the local min. you calculate the second order partial derivatives of $f(x,y)$ because it's a multivariate function you will calculate the Hessian matrix & its determinant to find out which of the two points is local min. & which is the local max.,more here: analyzemath.com/calculus/multivariable/maxima_minima.html
        $endgroup$
        – Blg Khalil
        Dec 3 '18 at 2:28


















      0












      $begingroup$

      We have,
      $$e^y = 2 lambda x$$
      $$x e^y = 2 lambda y$$
      thus,
      $$x = frac{2 lambda y}{e^y} = frac{2 lambda y}{2 lambda x} = frac{y}{x}$$
      finally,
      $$x^2 = y$$
      Now we replace into the original (constraint) equation
      $$y^2 + y - 6 = 0$$
      You have now the constraint in the form of a quadratic equation, you can easily find the value of $y$ through solving the equation.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thank you so much I just found the roots to be 2 and -3 but I'm confused what to do with those?
        $endgroup$
        – Emily
        Dec 2 '18 at 23:43










      • $begingroup$
        You repalce the values of $y$ that you have found in the constraint equation & you end up getting two points $(x_1,y_1)$ & $(x_2,y_2)$, one of these two points is the local max. while the other is the local min., to verify which one is the local max. & which one is the local min. you calculate the second order partial derivatives of $f(x,y)$ because it's a multivariate function you will calculate the Hessian matrix & its determinant to find out which of the two points is local min. & which is the local max.,more here: analyzemath.com/calculus/multivariable/maxima_minima.html
        $endgroup$
        – Blg Khalil
        Dec 3 '18 at 2:28
















      0












      0








      0





      $begingroup$

      We have,
      $$e^y = 2 lambda x$$
      $$x e^y = 2 lambda y$$
      thus,
      $$x = frac{2 lambda y}{e^y} = frac{2 lambda y}{2 lambda x} = frac{y}{x}$$
      finally,
      $$x^2 = y$$
      Now we replace into the original (constraint) equation
      $$y^2 + y - 6 = 0$$
      You have now the constraint in the form of a quadratic equation, you can easily find the value of $y$ through solving the equation.






      share|cite|improve this answer









      $endgroup$



      We have,
      $$e^y = 2 lambda x$$
      $$x e^y = 2 lambda y$$
      thus,
      $$x = frac{2 lambda y}{e^y} = frac{2 lambda y}{2 lambda x} = frac{y}{x}$$
      finally,
      $$x^2 = y$$
      Now we replace into the original (constraint) equation
      $$y^2 + y - 6 = 0$$
      You have now the constraint in the form of a quadratic equation, you can easily find the value of $y$ through solving the equation.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 2 '18 at 23:22









      Blg KhalilBlg Khalil

      283




      283












      • $begingroup$
        Thank you so much I just found the roots to be 2 and -3 but I'm confused what to do with those?
        $endgroup$
        – Emily
        Dec 2 '18 at 23:43










      • $begingroup$
        You repalce the values of $y$ that you have found in the constraint equation & you end up getting two points $(x_1,y_1)$ & $(x_2,y_2)$, one of these two points is the local max. while the other is the local min., to verify which one is the local max. & which one is the local min. you calculate the second order partial derivatives of $f(x,y)$ because it's a multivariate function you will calculate the Hessian matrix & its determinant to find out which of the two points is local min. & which is the local max.,more here: analyzemath.com/calculus/multivariable/maxima_minima.html
        $endgroup$
        – Blg Khalil
        Dec 3 '18 at 2:28




















      • $begingroup$
        Thank you so much I just found the roots to be 2 and -3 but I'm confused what to do with those?
        $endgroup$
        – Emily
        Dec 2 '18 at 23:43










      • $begingroup$
        You repalce the values of $y$ that you have found in the constraint equation & you end up getting two points $(x_1,y_1)$ & $(x_2,y_2)$, one of these two points is the local max. while the other is the local min., to verify which one is the local max. & which one is the local min. you calculate the second order partial derivatives of $f(x,y)$ because it's a multivariate function you will calculate the Hessian matrix & its determinant to find out which of the two points is local min. & which is the local max.,more here: analyzemath.com/calculus/multivariable/maxima_minima.html
        $endgroup$
        – Blg Khalil
        Dec 3 '18 at 2:28


















      $begingroup$
      Thank you so much I just found the roots to be 2 and -3 but I'm confused what to do with those?
      $endgroup$
      – Emily
      Dec 2 '18 at 23:43




      $begingroup$
      Thank you so much I just found the roots to be 2 and -3 but I'm confused what to do with those?
      $endgroup$
      – Emily
      Dec 2 '18 at 23:43












      $begingroup$
      You repalce the values of $y$ that you have found in the constraint equation & you end up getting two points $(x_1,y_1)$ & $(x_2,y_2)$, one of these two points is the local max. while the other is the local min., to verify which one is the local max. & which one is the local min. you calculate the second order partial derivatives of $f(x,y)$ because it's a multivariate function you will calculate the Hessian matrix & its determinant to find out which of the two points is local min. & which is the local max.,more here: analyzemath.com/calculus/multivariable/maxima_minima.html
      $endgroup$
      – Blg Khalil
      Dec 3 '18 at 2:28






      $begingroup$
      You repalce the values of $y$ that you have found in the constraint equation & you end up getting two points $(x_1,y_1)$ & $(x_2,y_2)$, one of these two points is the local max. while the other is the local min., to verify which one is the local max. & which one is the local min. you calculate the second order partial derivatives of $f(x,y)$ because it's a multivariate function you will calculate the Hessian matrix & its determinant to find out which of the two points is local min. & which is the local max.,more here: analyzemath.com/calculus/multivariable/maxima_minima.html
      $endgroup$
      – Blg Khalil
      Dec 3 '18 at 2:28













      0












      $begingroup$

      Following your calculations:
      $$2lambda=e^y/x$$
      $$2lambda=xe^y/y$$
      so
      $$e^y/x=xe^y/y iff ye^y=x^2e^y iff y=x^2$$ where in the last equivalency we use that $e^yne0$.



      Now, use this information with your constraint to get a bicuadratic equation on $x$ that you can solve alone.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Following your calculations:
        $$2lambda=e^y/x$$
        $$2lambda=xe^y/y$$
        so
        $$e^y/x=xe^y/y iff ye^y=x^2e^y iff y=x^2$$ where in the last equivalency we use that $e^yne0$.



        Now, use this information with your constraint to get a bicuadratic equation on $x$ that you can solve alone.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Following your calculations:
          $$2lambda=e^y/x$$
          $$2lambda=xe^y/y$$
          so
          $$e^y/x=xe^y/y iff ye^y=x^2e^y iff y=x^2$$ where in the last equivalency we use that $e^yne0$.



          Now, use this information with your constraint to get a bicuadratic equation on $x$ that you can solve alone.






          share|cite|improve this answer









          $endgroup$



          Following your calculations:
          $$2lambda=e^y/x$$
          $$2lambda=xe^y/y$$
          so
          $$e^y/x=xe^y/y iff ye^y=x^2e^y iff y=x^2$$ where in the last equivalency we use that $e^yne0$.



          Now, use this information with your constraint to get a bicuadratic equation on $x$ that you can solve alone.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 23:19









          Tito EliatronTito Eliatron

          1,446622




          1,446622























              0












              $begingroup$

              Without Lagrange Multipliers



              Calling



              $$
              x = rcos(theta)\
              y = rsin(theta)
              $$



              we have the equivalent problem



              $$
              maxmin_{theta}f(theta) = sqrt{6}cos(theta) e^{sqrt 6sin(theta)}
              $$



              and



              $$
              f'(theta) = sqrt{6}e^{sqrt{6} sin (theta )} left(sqrt{6} cos ^2(theta )-sin (theta )right)to sqrt{6} cos ^2(theta )-sin (theta ) = 0
              $$



              etc.



              NOTE



              This result is equivalent to the system



              $$
              x^2+y^2=6\
              x^2-y = 0
              $$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Without Lagrange Multipliers



                Calling



                $$
                x = rcos(theta)\
                y = rsin(theta)
                $$



                we have the equivalent problem



                $$
                maxmin_{theta}f(theta) = sqrt{6}cos(theta) e^{sqrt 6sin(theta)}
                $$



                and



                $$
                f'(theta) = sqrt{6}e^{sqrt{6} sin (theta )} left(sqrt{6} cos ^2(theta )-sin (theta )right)to sqrt{6} cos ^2(theta )-sin (theta ) = 0
                $$



                etc.



                NOTE



                This result is equivalent to the system



                $$
                x^2+y^2=6\
                x^2-y = 0
                $$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Without Lagrange Multipliers



                  Calling



                  $$
                  x = rcos(theta)\
                  y = rsin(theta)
                  $$



                  we have the equivalent problem



                  $$
                  maxmin_{theta}f(theta) = sqrt{6}cos(theta) e^{sqrt 6sin(theta)}
                  $$



                  and



                  $$
                  f'(theta) = sqrt{6}e^{sqrt{6} sin (theta )} left(sqrt{6} cos ^2(theta )-sin (theta )right)to sqrt{6} cos ^2(theta )-sin (theta ) = 0
                  $$



                  etc.



                  NOTE



                  This result is equivalent to the system



                  $$
                  x^2+y^2=6\
                  x^2-y = 0
                  $$






                  share|cite|improve this answer









                  $endgroup$



                  Without Lagrange Multipliers



                  Calling



                  $$
                  x = rcos(theta)\
                  y = rsin(theta)
                  $$



                  we have the equivalent problem



                  $$
                  maxmin_{theta}f(theta) = sqrt{6}cos(theta) e^{sqrt 6sin(theta)}
                  $$



                  and



                  $$
                  f'(theta) = sqrt{6}e^{sqrt{6} sin (theta )} left(sqrt{6} cos ^2(theta )-sin (theta )right)to sqrt{6} cos ^2(theta )-sin (theta ) = 0
                  $$



                  etc.



                  NOTE



                  This result is equivalent to the system



                  $$
                  x^2+y^2=6\
                  x^2-y = 0
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 3 '18 at 0:02









                  CesareoCesareo

                  8,6893516




                  8,6893516






























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                      Puebla de Zaragoza

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                      Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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                      Musa

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                      Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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