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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                      How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

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                      1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
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