Global extrema of $f(x,y)= e^{-4x^2-9y^2}(2x+3y)$ on the ellipse $4x^2+9y^2 leq 72$












1












$begingroup$


I'm asked to find the global extremas of the following function :$$f(x,y)= e^{-4x^2-9y^2}(2x+3y)$$ on the ellipse $4x^2+9y^2 leq 72$



So for the inside, I have $nabla f=(2e^{-4x^2-9y^2}+(-16x^2-24xy)e^{-4x^2-9y^2}, 3e^{-4x^2-9y^2}+(-36xy-54y^2)e^{-4x^2-9y^2})= (0, 0)$. I get two equations $$-16x^2-24xy+2=0$$ $$-54y^2-36xy+3=0$$ Adding $frac{-2}{3}$ of the second to the first gives $-16x^3+36y^2=0$,$to$ $9y^2=4x^2$ $to$ $ y = pm frac{2x}{3}$. Putting that in our original function, I get $e^{-8x^2}(4x)$ and $e^{-8x^2}(0)=0$
So deriving the first (because the second gives $0$) and setting it equal to $0$, I get $x=pm frac{1}{16}$ implies $y=pm frac{1}{24}$



For the border, I use Lagrange : $$nabla f = lambda nabla g$$ and get : $$-2 e^{-4 x^2 - 9 y^2} (8 x^2 + 12 x y - 1) = lambda 8x $$ and $$e^{-4 x^2 - 9 y^2} (-36 x y - 54 y^2 + 3)=lambda 18y$$
So $$frac{-2(8x^2+12xy-1)}{8x}= frac{(-36xy-54y^2+3)}{18y}$$
After rearranging, I get $-24x^2+12xy=-54y^2+1$ which seems a bit ugly. According to WolframAlpha, the solution for that is $y=frac{2x}{3}$ which is one of the value (the other being $-frac{2x}{3}) $ for the inside of the region that I got before, which I find a bit strange. So my questions are :


1) Is what I did until now even correct ?

2) If not, what did I do wrong and how should I proceed instead ?

3) If yes, how should I proceed now ?

4) Also for the inside values I got (i.e. $x=pm frac{1}{16}$, $y =pm frac{1}{24} $) : I just need to plug them into the function and at the end find the greatest and smallest values from all candidates, is that correct (I'm quite sure, just want to get a confirmation) ?



Thanks for your help !










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    I would start with replacing $hat x=2x$, $hat y=3y$ to get rid of all unnecessary coefficients. After that on the boundary: consider polar coordinates instead of Lagrange, i.e. parameterize the curve by cosine/sine.
    $endgroup$
    – A.Γ.
    Dec 17 '18 at 13:33












  • $begingroup$
    @A.Γ. So I would get $ e^{-r^2}(rcos(theta)+rsin(theta)$ Could I just plug in $r=6 sqrt{2}$ and then derivate with respect to $theta$ ?
    $endgroup$
    – Poujh
    Dec 17 '18 at 13:43








  • 1




    $begingroup$
    Right. $costheta+sintheta=sqrt{2}sin(theta+pi/4)$ can be also maximized without derivative.
    $endgroup$
    – A.Γ.
    Dec 17 '18 at 13:46








  • 2




    $begingroup$
    Just to check: change to the polar coordinates from the beginning. You will get the problem $$F(r,theta)=re^{-r^2}sqrt{2}sin(theta+pi/4),quad r^2le 72,thetain[0,2pi].$$ It is easy to optimize w.r.t. $theta$ (when $sin=pm 1$). Then differentiate w.r.t. $r$.
    $endgroup$
    – A.Γ.
    Dec 17 '18 at 14:07


















1












$begingroup$


I'm asked to find the global extremas of the following function :$$f(x,y)= e^{-4x^2-9y^2}(2x+3y)$$ on the ellipse $4x^2+9y^2 leq 72$



So for the inside, I have $nabla f=(2e^{-4x^2-9y^2}+(-16x^2-24xy)e^{-4x^2-9y^2}, 3e^{-4x^2-9y^2}+(-36xy-54y^2)e^{-4x^2-9y^2})= (0, 0)$. I get two equations $$-16x^2-24xy+2=0$$ $$-54y^2-36xy+3=0$$ Adding $frac{-2}{3}$ of the second to the first gives $-16x^3+36y^2=0$,$to$ $9y^2=4x^2$ $to$ $ y = pm frac{2x}{3}$. Putting that in our original function, I get $e^{-8x^2}(4x)$ and $e^{-8x^2}(0)=0$
So deriving the first (because the second gives $0$) and setting it equal to $0$, I get $x=pm frac{1}{16}$ implies $y=pm frac{1}{24}$



For the border, I use Lagrange : $$nabla f = lambda nabla g$$ and get : $$-2 e^{-4 x^2 - 9 y^2} (8 x^2 + 12 x y - 1) = lambda 8x $$ and $$e^{-4 x^2 - 9 y^2} (-36 x y - 54 y^2 + 3)=lambda 18y$$
So $$frac{-2(8x^2+12xy-1)}{8x}= frac{(-36xy-54y^2+3)}{18y}$$
After rearranging, I get $-24x^2+12xy=-54y^2+1$ which seems a bit ugly. According to WolframAlpha, the solution for that is $y=frac{2x}{3}$ which is one of the value (the other being $-frac{2x}{3}) $ for the inside of the region that I got before, which I find a bit strange. So my questions are :


1) Is what I did until now even correct ?

2) If not, what did I do wrong and how should I proceed instead ?

3) If yes, how should I proceed now ?

4) Also for the inside values I got (i.e. $x=pm frac{1}{16}$, $y =pm frac{1}{24} $) : I just need to plug them into the function and at the end find the greatest and smallest values from all candidates, is that correct (I'm quite sure, just want to get a confirmation) ?



Thanks for your help !










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    I would start with replacing $hat x=2x$, $hat y=3y$ to get rid of all unnecessary coefficients. After that on the boundary: consider polar coordinates instead of Lagrange, i.e. parameterize the curve by cosine/sine.
    $endgroup$
    – A.Γ.
    Dec 17 '18 at 13:33












  • $begingroup$
    @A.Γ. So I would get $ e^{-r^2}(rcos(theta)+rsin(theta)$ Could I just plug in $r=6 sqrt{2}$ and then derivate with respect to $theta$ ?
    $endgroup$
    – Poujh
    Dec 17 '18 at 13:43








  • 1




    $begingroup$
    Right. $costheta+sintheta=sqrt{2}sin(theta+pi/4)$ can be also maximized without derivative.
    $endgroup$
    – A.Γ.
    Dec 17 '18 at 13:46








  • 2




    $begingroup$
    Just to check: change to the polar coordinates from the beginning. You will get the problem $$F(r,theta)=re^{-r^2}sqrt{2}sin(theta+pi/4),quad r^2le 72,thetain[0,2pi].$$ It is easy to optimize w.r.t. $theta$ (when $sin=pm 1$). Then differentiate w.r.t. $r$.
    $endgroup$
    – A.Γ.
    Dec 17 '18 at 14:07
















1












1








1





$begingroup$


I'm asked to find the global extremas of the following function :$$f(x,y)= e^{-4x^2-9y^2}(2x+3y)$$ on the ellipse $4x^2+9y^2 leq 72$



So for the inside, I have $nabla f=(2e^{-4x^2-9y^2}+(-16x^2-24xy)e^{-4x^2-9y^2}, 3e^{-4x^2-9y^2}+(-36xy-54y^2)e^{-4x^2-9y^2})= (0, 0)$. I get two equations $$-16x^2-24xy+2=0$$ $$-54y^2-36xy+3=0$$ Adding $frac{-2}{3}$ of the second to the first gives $-16x^3+36y^2=0$,$to$ $9y^2=4x^2$ $to$ $ y = pm frac{2x}{3}$. Putting that in our original function, I get $e^{-8x^2}(4x)$ and $e^{-8x^2}(0)=0$
So deriving the first (because the second gives $0$) and setting it equal to $0$, I get $x=pm frac{1}{16}$ implies $y=pm frac{1}{24}$



For the border, I use Lagrange : $$nabla f = lambda nabla g$$ and get : $$-2 e^{-4 x^2 - 9 y^2} (8 x^2 + 12 x y - 1) = lambda 8x $$ and $$e^{-4 x^2 - 9 y^2} (-36 x y - 54 y^2 + 3)=lambda 18y$$
So $$frac{-2(8x^2+12xy-1)}{8x}= frac{(-36xy-54y^2+3)}{18y}$$
After rearranging, I get $-24x^2+12xy=-54y^2+1$ which seems a bit ugly. According to WolframAlpha, the solution for that is $y=frac{2x}{3}$ which is one of the value (the other being $-frac{2x}{3}) $ for the inside of the region that I got before, which I find a bit strange. So my questions are :


1) Is what I did until now even correct ?

2) If not, what did I do wrong and how should I proceed instead ?

3) If yes, how should I proceed now ?

4) Also for the inside values I got (i.e. $x=pm frac{1}{16}$, $y =pm frac{1}{24} $) : I just need to plug them into the function and at the end find the greatest and smallest values from all candidates, is that correct (I'm quite sure, just want to get a confirmation) ?



Thanks for your help !










share|cite|improve this question











$endgroup$




I'm asked to find the global extremas of the following function :$$f(x,y)= e^{-4x^2-9y^2}(2x+3y)$$ on the ellipse $4x^2+9y^2 leq 72$



So for the inside, I have $nabla f=(2e^{-4x^2-9y^2}+(-16x^2-24xy)e^{-4x^2-9y^2}, 3e^{-4x^2-9y^2}+(-36xy-54y^2)e^{-4x^2-9y^2})= (0, 0)$. I get two equations $$-16x^2-24xy+2=0$$ $$-54y^2-36xy+3=0$$ Adding $frac{-2}{3}$ of the second to the first gives $-16x^3+36y^2=0$,$to$ $9y^2=4x^2$ $to$ $ y = pm frac{2x}{3}$. Putting that in our original function, I get $e^{-8x^2}(4x)$ and $e^{-8x^2}(0)=0$
So deriving the first (because the second gives $0$) and setting it equal to $0$, I get $x=pm frac{1}{16}$ implies $y=pm frac{1}{24}$



For the border, I use Lagrange : $$nabla f = lambda nabla g$$ and get : $$-2 e^{-4 x^2 - 9 y^2} (8 x^2 + 12 x y - 1) = lambda 8x $$ and $$e^{-4 x^2 - 9 y^2} (-36 x y - 54 y^2 + 3)=lambda 18y$$
So $$frac{-2(8x^2+12xy-1)}{8x}= frac{(-36xy-54y^2+3)}{18y}$$
After rearranging, I get $-24x^2+12xy=-54y^2+1$ which seems a bit ugly. According to WolframAlpha, the solution for that is $y=frac{2x}{3}$ which is one of the value (the other being $-frac{2x}{3}) $ for the inside of the region that I got before, which I find a bit strange. So my questions are :


1) Is what I did until now even correct ?

2) If not, what did I do wrong and how should I proceed instead ?

3) If yes, how should I proceed now ?

4) Also for the inside values I got (i.e. $x=pm frac{1}{16}$, $y =pm frac{1}{24} $) : I just need to plug them into the function and at the end find the greatest and smallest values from all candidates, is that correct (I'm quite sure, just want to get a confirmation) ?



Thanks for your help !







real-analysis calculus analysis multivariable-calculus lagrange-multiplier






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share|cite|improve this question













share|cite|improve this question




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edited Jan 7 at 19:25







Poujh

















asked Dec 17 '18 at 13:09









PoujhPoujh

6111516




6111516








  • 3




    $begingroup$
    I would start with replacing $hat x=2x$, $hat y=3y$ to get rid of all unnecessary coefficients. After that on the boundary: consider polar coordinates instead of Lagrange, i.e. parameterize the curve by cosine/sine.
    $endgroup$
    – A.Γ.
    Dec 17 '18 at 13:33












  • $begingroup$
    @A.Γ. So I would get $ e^{-r^2}(rcos(theta)+rsin(theta)$ Could I just plug in $r=6 sqrt{2}$ and then derivate with respect to $theta$ ?
    $endgroup$
    – Poujh
    Dec 17 '18 at 13:43








  • 1




    $begingroup$
    Right. $costheta+sintheta=sqrt{2}sin(theta+pi/4)$ can be also maximized without derivative.
    $endgroup$
    – A.Γ.
    Dec 17 '18 at 13:46








  • 2




    $begingroup$
    Just to check: change to the polar coordinates from the beginning. You will get the problem $$F(r,theta)=re^{-r^2}sqrt{2}sin(theta+pi/4),quad r^2le 72,thetain[0,2pi].$$ It is easy to optimize w.r.t. $theta$ (when $sin=pm 1$). Then differentiate w.r.t. $r$.
    $endgroup$
    – A.Γ.
    Dec 17 '18 at 14:07
















  • 3




    $begingroup$
    I would start with replacing $hat x=2x$, $hat y=3y$ to get rid of all unnecessary coefficients. After that on the boundary: consider polar coordinates instead of Lagrange, i.e. parameterize the curve by cosine/sine.
    $endgroup$
    – A.Γ.
    Dec 17 '18 at 13:33












  • $begingroup$
    @A.Γ. So I would get $ e^{-r^2}(rcos(theta)+rsin(theta)$ Could I just plug in $r=6 sqrt{2}$ and then derivate with respect to $theta$ ?
    $endgroup$
    – Poujh
    Dec 17 '18 at 13:43








  • 1




    $begingroup$
    Right. $costheta+sintheta=sqrt{2}sin(theta+pi/4)$ can be also maximized without derivative.
    $endgroup$
    – A.Γ.
    Dec 17 '18 at 13:46








  • 2




    $begingroup$
    Just to check: change to the polar coordinates from the beginning. You will get the problem $$F(r,theta)=re^{-r^2}sqrt{2}sin(theta+pi/4),quad r^2le 72,thetain[0,2pi].$$ It is easy to optimize w.r.t. $theta$ (when $sin=pm 1$). Then differentiate w.r.t. $r$.
    $endgroup$
    – A.Γ.
    Dec 17 '18 at 14:07










3




3




$begingroup$
I would start with replacing $hat x=2x$, $hat y=3y$ to get rid of all unnecessary coefficients. After that on the boundary: consider polar coordinates instead of Lagrange, i.e. parameterize the curve by cosine/sine.
$endgroup$
– A.Γ.
Dec 17 '18 at 13:33






$begingroup$
I would start with replacing $hat x=2x$, $hat y=3y$ to get rid of all unnecessary coefficients. After that on the boundary: consider polar coordinates instead of Lagrange, i.e. parameterize the curve by cosine/sine.
$endgroup$
– A.Γ.
Dec 17 '18 at 13:33














$begingroup$
@A.Γ. So I would get $ e^{-r^2}(rcos(theta)+rsin(theta)$ Could I just plug in $r=6 sqrt{2}$ and then derivate with respect to $theta$ ?
$endgroup$
– Poujh
Dec 17 '18 at 13:43






$begingroup$
@A.Γ. So I would get $ e^{-r^2}(rcos(theta)+rsin(theta)$ Could I just plug in $r=6 sqrt{2}$ and then derivate with respect to $theta$ ?
$endgroup$
– Poujh
Dec 17 '18 at 13:43






1




1




$begingroup$
Right. $costheta+sintheta=sqrt{2}sin(theta+pi/4)$ can be also maximized without derivative.
$endgroup$
– A.Γ.
Dec 17 '18 at 13:46






$begingroup$
Right. $costheta+sintheta=sqrt{2}sin(theta+pi/4)$ can be also maximized without derivative.
$endgroup$
– A.Γ.
Dec 17 '18 at 13:46






2




2




$begingroup$
Just to check: change to the polar coordinates from the beginning. You will get the problem $$F(r,theta)=re^{-r^2}sqrt{2}sin(theta+pi/4),quad r^2le 72,thetain[0,2pi].$$ It is easy to optimize w.r.t. $theta$ (when $sin=pm 1$). Then differentiate w.r.t. $r$.
$endgroup$
– A.Γ.
Dec 17 '18 at 14:07






$begingroup$
Just to check: change to the polar coordinates from the beginning. You will get the problem $$F(r,theta)=re^{-r^2}sqrt{2}sin(theta+pi/4),quad r^2le 72,thetain[0,2pi].$$ It is easy to optimize w.r.t. $theta$ (when $sin=pm 1$). Then differentiate w.r.t. $r$.
$endgroup$
– A.Γ.
Dec 17 '18 at 14:07












2 Answers
2






active

oldest

votes


















2












$begingroup$

From the lagrangian



$$
L(x,y,mu,epsilon) = f(x,y) + mu(g(x,y)-72+epsilon^2)
$$



here $epsilon$ is a slack variable to transform the inequality constraint into an equality.



The stationary conditions are



$$
left{
begin{array}{rcl}
8 lambda x-8 (2 x+3 y) x+2& = & 0 \
18 lambda y-18 (2 x+3 y) y+3& =& 0 \
epsilon ^2+4 x^2+9 y^2-72 & = & 0 \
2 epsilon lambda & = & 0 \
end{array}
right.
$$



here $lambda = mu e^{4x^2+9y^2}$. After solving we obtain



$$
begin{array}{ccccc}
x & y & lambda & epsilon & f(x,y) \
-3 & -2 & -frac{143}{12} & 0 & -frac{12}{e^{72}} \
-frac{1}{4} & -frac{1}{6} & 0 & sqrt{frac{143}{2}} & -frac{1}{sqrt{e}} \
frac{1}{4} & frac{1}{6} & 0 & sqrt{frac{143}{2}} & frac{1}{sqrt{e}} \
3 & 2 & frac{143}{12} & 0 & frac{12}{e^{72}} \
end{array}
$$



NOTE



Solutions with $epsilon = 0$ are solutions at the boundary and solutions with $epsilon ne 0$ are interior at the feasible region.



Attached two plots. The first shows the location for the solution points and the second shows a detail for the two internal solutions.



enter image description here



enter image description here






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your answer: I tried to solve the problem manually, that's why I didn't respond immediately to your answer (because your method obviously needs a computer). I was finally able to come by hand on the same values of x and y as you got using A.Γ. hints. Thank you !
    $endgroup$
    – Poujh
    Dec 17 '18 at 17:36



















1












$begingroup$

Okay, with the hints, I will try to answer my own question, maybe this will help someone in the future looking at this question.



If we substitute $hat{x}=2x$ and $hat{y}=3y$, we get $e^{-hat{x}-hat{y}}(hat{x}+hat{y})$ Now, if we take the partial derivatives of that, set them equal to 0 and solve, we get finally $hat{y}=pm hat{x}$. Now, if we subtitute that into our original function and take the derivative, we get $hat{x}=frac{1}{2}$, so $x=frac{1}{4}$
$hat{y}=pm hat{x}$ can be rewritten as $y=pm frac{2}{3}x$. So now we get $y=frac{1}{6}$ and $y=frac{-1}{6}$ as in Cesareo's answer.



On the rand, if we parametrize using polar coordinates with $r=sqrt{72}=6sqrt{2}$,
$hat{x}=rcos(theta), hat{y}=rsin(theta)$, plug that into our function, take the derivative (now in polar coordinates) with respect to $theta$ (remember that $r$ is obviously fixed) and set it equal to zero, we get $theta = frac{pi}{4}$ and $theta = frac{5 pi}{4}$. So $hat{x}= r cos(theta)$ and $hat{y}=rsin(theta)$. So we get $hat{x}=pm6=hat{y}$. So we get $x=frac{hat{x}}{2}=pm3$ and $y=frac{hat{y}}{3}=pm2$, again as in Cesareo's answer.



Now, we would just have to plug our different values into our function to see where the global maximum/ minimum is.



Maybe this will help someone in the future.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    From the lagrangian



    $$
    L(x,y,mu,epsilon) = f(x,y) + mu(g(x,y)-72+epsilon^2)
    $$



    here $epsilon$ is a slack variable to transform the inequality constraint into an equality.



    The stationary conditions are



    $$
    left{
    begin{array}{rcl}
    8 lambda x-8 (2 x+3 y) x+2& = & 0 \
    18 lambda y-18 (2 x+3 y) y+3& =& 0 \
    epsilon ^2+4 x^2+9 y^2-72 & = & 0 \
    2 epsilon lambda & = & 0 \
    end{array}
    right.
    $$



    here $lambda = mu e^{4x^2+9y^2}$. After solving we obtain



    $$
    begin{array}{ccccc}
    x & y & lambda & epsilon & f(x,y) \
    -3 & -2 & -frac{143}{12} & 0 & -frac{12}{e^{72}} \
    -frac{1}{4} & -frac{1}{6} & 0 & sqrt{frac{143}{2}} & -frac{1}{sqrt{e}} \
    frac{1}{4} & frac{1}{6} & 0 & sqrt{frac{143}{2}} & frac{1}{sqrt{e}} \
    3 & 2 & frac{143}{12} & 0 & frac{12}{e^{72}} \
    end{array}
    $$



    NOTE



    Solutions with $epsilon = 0$ are solutions at the boundary and solutions with $epsilon ne 0$ are interior at the feasible region.



    Attached two plots. The first shows the location for the solution points and the second shows a detail for the two internal solutions.



    enter image description here



    enter image description here






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for your answer: I tried to solve the problem manually, that's why I didn't respond immediately to your answer (because your method obviously needs a computer). I was finally able to come by hand on the same values of x and y as you got using A.Γ. hints. Thank you !
      $endgroup$
      – Poujh
      Dec 17 '18 at 17:36
















    2












    $begingroup$

    From the lagrangian



    $$
    L(x,y,mu,epsilon) = f(x,y) + mu(g(x,y)-72+epsilon^2)
    $$



    here $epsilon$ is a slack variable to transform the inequality constraint into an equality.



    The stationary conditions are



    $$
    left{
    begin{array}{rcl}
    8 lambda x-8 (2 x+3 y) x+2& = & 0 \
    18 lambda y-18 (2 x+3 y) y+3& =& 0 \
    epsilon ^2+4 x^2+9 y^2-72 & = & 0 \
    2 epsilon lambda & = & 0 \
    end{array}
    right.
    $$



    here $lambda = mu e^{4x^2+9y^2}$. After solving we obtain



    $$
    begin{array}{ccccc}
    x & y & lambda & epsilon & f(x,y) \
    -3 & -2 & -frac{143}{12} & 0 & -frac{12}{e^{72}} \
    -frac{1}{4} & -frac{1}{6} & 0 & sqrt{frac{143}{2}} & -frac{1}{sqrt{e}} \
    frac{1}{4} & frac{1}{6} & 0 & sqrt{frac{143}{2}} & frac{1}{sqrt{e}} \
    3 & 2 & frac{143}{12} & 0 & frac{12}{e^{72}} \
    end{array}
    $$



    NOTE



    Solutions with $epsilon = 0$ are solutions at the boundary and solutions with $epsilon ne 0$ are interior at the feasible region.



    Attached two plots. The first shows the location for the solution points and the second shows a detail for the two internal solutions.



    enter image description here



    enter image description here






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for your answer: I tried to solve the problem manually, that's why I didn't respond immediately to your answer (because your method obviously needs a computer). I was finally able to come by hand on the same values of x and y as you got using A.Γ. hints. Thank you !
      $endgroup$
      – Poujh
      Dec 17 '18 at 17:36














    2












    2








    2





    $begingroup$

    From the lagrangian



    $$
    L(x,y,mu,epsilon) = f(x,y) + mu(g(x,y)-72+epsilon^2)
    $$



    here $epsilon$ is a slack variable to transform the inequality constraint into an equality.



    The stationary conditions are



    $$
    left{
    begin{array}{rcl}
    8 lambda x-8 (2 x+3 y) x+2& = & 0 \
    18 lambda y-18 (2 x+3 y) y+3& =& 0 \
    epsilon ^2+4 x^2+9 y^2-72 & = & 0 \
    2 epsilon lambda & = & 0 \
    end{array}
    right.
    $$



    here $lambda = mu e^{4x^2+9y^2}$. After solving we obtain



    $$
    begin{array}{ccccc}
    x & y & lambda & epsilon & f(x,y) \
    -3 & -2 & -frac{143}{12} & 0 & -frac{12}{e^{72}} \
    -frac{1}{4} & -frac{1}{6} & 0 & sqrt{frac{143}{2}} & -frac{1}{sqrt{e}} \
    frac{1}{4} & frac{1}{6} & 0 & sqrt{frac{143}{2}} & frac{1}{sqrt{e}} \
    3 & 2 & frac{143}{12} & 0 & frac{12}{e^{72}} \
    end{array}
    $$



    NOTE



    Solutions with $epsilon = 0$ are solutions at the boundary and solutions with $epsilon ne 0$ are interior at the feasible region.



    Attached two plots. The first shows the location for the solution points and the second shows a detail for the two internal solutions.



    enter image description here



    enter image description here






    share|cite|improve this answer









    $endgroup$



    From the lagrangian



    $$
    L(x,y,mu,epsilon) = f(x,y) + mu(g(x,y)-72+epsilon^2)
    $$



    here $epsilon$ is a slack variable to transform the inequality constraint into an equality.



    The stationary conditions are



    $$
    left{
    begin{array}{rcl}
    8 lambda x-8 (2 x+3 y) x+2& = & 0 \
    18 lambda y-18 (2 x+3 y) y+3& =& 0 \
    epsilon ^2+4 x^2+9 y^2-72 & = & 0 \
    2 epsilon lambda & = & 0 \
    end{array}
    right.
    $$



    here $lambda = mu e^{4x^2+9y^2}$. After solving we obtain



    $$
    begin{array}{ccccc}
    x & y & lambda & epsilon & f(x,y) \
    -3 & -2 & -frac{143}{12} & 0 & -frac{12}{e^{72}} \
    -frac{1}{4} & -frac{1}{6} & 0 & sqrt{frac{143}{2}} & -frac{1}{sqrt{e}} \
    frac{1}{4} & frac{1}{6} & 0 & sqrt{frac{143}{2}} & frac{1}{sqrt{e}} \
    3 & 2 & frac{143}{12} & 0 & frac{12}{e^{72}} \
    end{array}
    $$



    NOTE



    Solutions with $epsilon = 0$ are solutions at the boundary and solutions with $epsilon ne 0$ are interior at the feasible region.



    Attached two plots. The first shows the location for the solution points and the second shows a detail for the two internal solutions.



    enter image description here



    enter image description here







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 17 '18 at 14:24









    CesareoCesareo

    9,8463517




    9,8463517












    • $begingroup$
      Thanks for your answer: I tried to solve the problem manually, that's why I didn't respond immediately to your answer (because your method obviously needs a computer). I was finally able to come by hand on the same values of x and y as you got using A.Γ. hints. Thank you !
      $endgroup$
      – Poujh
      Dec 17 '18 at 17:36


















    • $begingroup$
      Thanks for your answer: I tried to solve the problem manually, that's why I didn't respond immediately to your answer (because your method obviously needs a computer). I was finally able to come by hand on the same values of x and y as you got using A.Γ. hints. Thank you !
      $endgroup$
      – Poujh
      Dec 17 '18 at 17:36
















    $begingroup$
    Thanks for your answer: I tried to solve the problem manually, that's why I didn't respond immediately to your answer (because your method obviously needs a computer). I was finally able to come by hand on the same values of x and y as you got using A.Γ. hints. Thank you !
    $endgroup$
    – Poujh
    Dec 17 '18 at 17:36




    $begingroup$
    Thanks for your answer: I tried to solve the problem manually, that's why I didn't respond immediately to your answer (because your method obviously needs a computer). I was finally able to come by hand on the same values of x and y as you got using A.Γ. hints. Thank you !
    $endgroup$
    – Poujh
    Dec 17 '18 at 17:36











    1












    $begingroup$

    Okay, with the hints, I will try to answer my own question, maybe this will help someone in the future looking at this question.



    If we substitute $hat{x}=2x$ and $hat{y}=3y$, we get $e^{-hat{x}-hat{y}}(hat{x}+hat{y})$ Now, if we take the partial derivatives of that, set them equal to 0 and solve, we get finally $hat{y}=pm hat{x}$. Now, if we subtitute that into our original function and take the derivative, we get $hat{x}=frac{1}{2}$, so $x=frac{1}{4}$
    $hat{y}=pm hat{x}$ can be rewritten as $y=pm frac{2}{3}x$. So now we get $y=frac{1}{6}$ and $y=frac{-1}{6}$ as in Cesareo's answer.



    On the rand, if we parametrize using polar coordinates with $r=sqrt{72}=6sqrt{2}$,
    $hat{x}=rcos(theta), hat{y}=rsin(theta)$, plug that into our function, take the derivative (now in polar coordinates) with respect to $theta$ (remember that $r$ is obviously fixed) and set it equal to zero, we get $theta = frac{pi}{4}$ and $theta = frac{5 pi}{4}$. So $hat{x}= r cos(theta)$ and $hat{y}=rsin(theta)$. So we get $hat{x}=pm6=hat{y}$. So we get $x=frac{hat{x}}{2}=pm3$ and $y=frac{hat{y}}{3}=pm2$, again as in Cesareo's answer.



    Now, we would just have to plug our different values into our function to see where the global maximum/ minimum is.



    Maybe this will help someone in the future.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Okay, with the hints, I will try to answer my own question, maybe this will help someone in the future looking at this question.



      If we substitute $hat{x}=2x$ and $hat{y}=3y$, we get $e^{-hat{x}-hat{y}}(hat{x}+hat{y})$ Now, if we take the partial derivatives of that, set them equal to 0 and solve, we get finally $hat{y}=pm hat{x}$. Now, if we subtitute that into our original function and take the derivative, we get $hat{x}=frac{1}{2}$, so $x=frac{1}{4}$
      $hat{y}=pm hat{x}$ can be rewritten as $y=pm frac{2}{3}x$. So now we get $y=frac{1}{6}$ and $y=frac{-1}{6}$ as in Cesareo's answer.



      On the rand, if we parametrize using polar coordinates with $r=sqrt{72}=6sqrt{2}$,
      $hat{x}=rcos(theta), hat{y}=rsin(theta)$, plug that into our function, take the derivative (now in polar coordinates) with respect to $theta$ (remember that $r$ is obviously fixed) and set it equal to zero, we get $theta = frac{pi}{4}$ and $theta = frac{5 pi}{4}$. So $hat{x}= r cos(theta)$ and $hat{y}=rsin(theta)$. So we get $hat{x}=pm6=hat{y}$. So we get $x=frac{hat{x}}{2}=pm3$ and $y=frac{hat{y}}{3}=pm2$, again as in Cesareo's answer.



      Now, we would just have to plug our different values into our function to see where the global maximum/ minimum is.



      Maybe this will help someone in the future.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Okay, with the hints, I will try to answer my own question, maybe this will help someone in the future looking at this question.



        If we substitute $hat{x}=2x$ and $hat{y}=3y$, we get $e^{-hat{x}-hat{y}}(hat{x}+hat{y})$ Now, if we take the partial derivatives of that, set them equal to 0 and solve, we get finally $hat{y}=pm hat{x}$. Now, if we subtitute that into our original function and take the derivative, we get $hat{x}=frac{1}{2}$, so $x=frac{1}{4}$
        $hat{y}=pm hat{x}$ can be rewritten as $y=pm frac{2}{3}x$. So now we get $y=frac{1}{6}$ and $y=frac{-1}{6}$ as in Cesareo's answer.



        On the rand, if we parametrize using polar coordinates with $r=sqrt{72}=6sqrt{2}$,
        $hat{x}=rcos(theta), hat{y}=rsin(theta)$, plug that into our function, take the derivative (now in polar coordinates) with respect to $theta$ (remember that $r$ is obviously fixed) and set it equal to zero, we get $theta = frac{pi}{4}$ and $theta = frac{5 pi}{4}$. So $hat{x}= r cos(theta)$ and $hat{y}=rsin(theta)$. So we get $hat{x}=pm6=hat{y}$. So we get $x=frac{hat{x}}{2}=pm3$ and $y=frac{hat{y}}{3}=pm2$, again as in Cesareo's answer.



        Now, we would just have to plug our different values into our function to see where the global maximum/ minimum is.



        Maybe this will help someone in the future.






        share|cite|improve this answer









        $endgroup$



        Okay, with the hints, I will try to answer my own question, maybe this will help someone in the future looking at this question.



        If we substitute $hat{x}=2x$ and $hat{y}=3y$, we get $e^{-hat{x}-hat{y}}(hat{x}+hat{y})$ Now, if we take the partial derivatives of that, set them equal to 0 and solve, we get finally $hat{y}=pm hat{x}$. Now, if we subtitute that into our original function and take the derivative, we get $hat{x}=frac{1}{2}$, so $x=frac{1}{4}$
        $hat{y}=pm hat{x}$ can be rewritten as $y=pm frac{2}{3}x$. So now we get $y=frac{1}{6}$ and $y=frac{-1}{6}$ as in Cesareo's answer.



        On the rand, if we parametrize using polar coordinates with $r=sqrt{72}=6sqrt{2}$,
        $hat{x}=rcos(theta), hat{y}=rsin(theta)$, plug that into our function, take the derivative (now in polar coordinates) with respect to $theta$ (remember that $r$ is obviously fixed) and set it equal to zero, we get $theta = frac{pi}{4}$ and $theta = frac{5 pi}{4}$. So $hat{x}= r cos(theta)$ and $hat{y}=rsin(theta)$. So we get $hat{x}=pm6=hat{y}$. So we get $x=frac{hat{x}}{2}=pm3$ and $y=frac{hat{y}}{3}=pm2$, again as in Cesareo's answer.



        Now, we would just have to plug our different values into our function to see where the global maximum/ minimum is.



        Maybe this will help someone in the future.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 17 '18 at 22:18









        PoujhPoujh

        6111516




        6111516






























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