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 March 8th, 2014, 07:13 AM #1 Newbie   Joined: Oct 2013 Posts: 29 Thanks: 1 Extrema of multivariable functions - systems of equations Hello, I have been having some problems with extrema of multivariable functions, but not with the algorithm itself, but mostly with the difficult systems of equations that arise from setting gradient of the function equal to zero. At the moment, I am struggling with the following two problems: 1) Find the extrema of $f(x,y)= xy\ln (x^{2}+y^{2})$ This leads to the system of equations: $y(ln(x^{2}+y^{2})+\frac{2x^{2}}{x^{2}+y^{2}})=0$ $x(ln(x^{2}+y^{2})+\frac{2y^{2}}{x^{2}+y^{2}})=0$ This would be easy if y and x were nonzero, I would divide the first equation by y, second by x and then subtract one from another. Sadly this is not the case. 2) Find the extrema of $f(x,y)=x^{2}+y^{2}+xy-4\ln x-10\ln y; x,y>0=$ The system of equations i get here is: $2x+y-\frac{4}{x}=0$ $2y+x-\frac{10}{y}=0$ It looks simple at first, but I always manage to arrive at a high order polynomial and never manage to get the actual coordinates of the stationary points. Maybe I am making an elementary mistake or missing something. Any input would be greatly appreciated.
 March 8th, 2014, 08:57 AM #2 Newbie   Joined: Mar 2014 Posts: 9 Thanks: 0 Re: Extrema of multivariable functions - systems of equation I don't know what's up with the first problem, but in the second wouldn't you get four x-coordinates? Make y subject in the first equation and put it into the second. You'd end up with: 3x^4 - 19x^2 + 16 = 0 Using quadratic formula you'd get: x^2 = (19+13)/6 x^2 = (19-13)/6
 March 8th, 2014, 09:40 AM #3 Newbie   Joined: Mar 2014 Posts: 9 Thanks: 0 Re: Extrema of multivariable functions - systems of equation All right! So finally got it for the first problem. Use polar coordinates: x=rcos? y=rsin? The equations become: rsin?((ln r^2) + 2(cos?)^2)=0 rcos?((lnr^2) + 2(sin?)^2)=0 2(rln r)sin? + 2rsin?(cos?)^2=0 2(rln r)cos? + 2rcos?(sin?)^2=0 Make 2rln r subjects in both and equate: -2r(cos?)^2=-2r(sin?)^2 cos?=sin? ?=?/4 Substitute back into 2(rln r)sin? + 2rsin?(cos?)^2=0 : ?2 rln r + r?2/2 = 0 ln r + 0.5 = 0 r=e^-0.5 Substitute r and ? back into x=rcos? and y=rsin? x=y=(1/?2)e^(-0.5)
 March 8th, 2014, 10:00 AM #4 Newbie   Joined: Oct 2013 Posts: 29 Thanks: 1 Re: Extrema of multivariable functions - systems of equation Thanks man, I have just finished the first one - haven't thought of using the quadratic formula like that. I will get the second one done now.
 March 8th, 2014, 11:09 AM #5 Newbie   Joined: Oct 2013 Posts: 29 Thanks: 1 Re: Extrema of multivariable functions - systems of equation Just arrived at the same result with your instructions. Have to say, truly tremendous work - I would have never managed myself. Just one more question. The answer from the textbook suggest that it is not $x=y$, but $x=\pm y$. Any idea how it could be deduced?
 March 8th, 2014, 11:28 AM #6 Newbie   Joined: Mar 2014 Posts: 9 Thanks: 0 Re: Extrema of multivariable functions - systems of equation Well... The final answer involves square roots, so it makes sense that it should have plus/minus. x=(1/?2)e^(-0.5) and y=(1/?2)e^(-0.5) OR x=(1/?2)e^(-0.5) and y=(-1/?2)e^(-0.5) OR x=(-1/?2)e^(-0.5) and y=(1/?2)e^(-0.5) OR x=(-1/?2)e^(-0.5) and y=(-1/?2)e^(-0.5) Two of them are just repetitions, so this reduces to: x=(1/?2)e^(-0.5) and y=(1/?2)e^(-0.5) OR x=(-1/?2)e^(-0.5) and y=(1/?2)e^(-0.5) So.. x=±y

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