Maxima inside open set












0












$begingroup$


Consider the function



$$f(x,y) = e^xcos(y).$$



I am asked if this function has a maximum or a minimum inside the unit circle ${x,y: x^2+y^2<1}$.



The answer provided here is that, since the partial derivatives of the function are never $0$ simultaneously, the gradient is never zero, so the function has no minima or maxima. The answer is not in English so it would be difficult to quote it exactly here, but the wording implies there is no maxima or minima in any set.



But let's say we included the points ${x,y:x^2+y^2=1}$. Wouldn't the function now have a maxima at the "endpoints" $x^2+y^2=1$? To me it seems there is probably a miswording, and the author meant that the function has no maxima on the open set in question, as the function grows arbitrarily large as we approach the edge of the set (while never reaching $x^2+y^2=1$). Or have I misunderstood something?






multivariable-calculus maxima-minima







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$endgroup$








  • 1




    $begingroup$
    Note that the function is harmonic
    $endgroup$
    – Gabriele Cassese
    Jan 28 at 18:06










  • $begingroup$
    @gabrielecassese Did you see this just by looking at the function ? Or you went through the calculations ?
    $endgroup$
    – Thinking
    Jan 29 at 17:57










  • $begingroup$
    This kind of function ($e^xsin(y), e^ycos(x)$) is a quite often used example of harmonic function.
    $endgroup$
    – Gabriele Cassese
    Jan 29 at 18:09
















0












$begingroup$


Consider the function



$$f(x,y) = e^xcos(y).$$



I am asked if this function has a maximum or a minimum inside the unit circle ${x,y: x^2+y^2<1}$.



The answer provided here is that, since the partial derivatives of the function are never $0$ simultaneously, the gradient is never zero, so the function has no minima or maxima. The answer is not in English so it would be difficult to quote it exactly here, but the wording implies there is no maxima or minima in any set.



But let's say we included the points ${x,y:x^2+y^2=1}$. Wouldn't the function now have a maxima at the "endpoints" $x^2+y^2=1$? To me it seems there is probably a miswording, and the author meant that the function has no maxima on the open set in question, as the function grows arbitrarily large as we approach the edge of the set (while never reaching $x^2+y^2=1$). Or have I misunderstood something?






multivariable-calculus maxima-minima







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Note that the function is harmonic
    $endgroup$
    – Gabriele Cassese
    Jan 28 at 18:06










  • $begingroup$
    @gabrielecassese Did you see this just by looking at the function ? Or you went through the calculations ?
    $endgroup$
    – Thinking
    Jan 29 at 17:57










  • $begingroup$
    This kind of function ($e^xsin(y), e^ycos(x)$) is a quite often used example of harmonic function.
    $endgroup$
    – Gabriele Cassese
    Jan 29 at 18:09














0












0








0





$begingroup$


Consider the function



$$f(x,y) = e^xcos(y).$$



I am asked if this function has a maximum or a minimum inside the unit circle ${x,y: x^2+y^2<1}$.



The answer provided here is that, since the partial derivatives of the function are never $0$ simultaneously, the gradient is never zero, so the function has no minima or maxima. The answer is not in English so it would be difficult to quote it exactly here, but the wording implies there is no maxima or minima in any set.



But let's say we included the points ${x,y:x^2+y^2=1}$. Wouldn't the function now have a maxima at the "endpoints" $x^2+y^2=1$? To me it seems there is probably a miswording, and the author meant that the function has no maxima on the open set in question, as the function grows arbitrarily large as we approach the edge of the set (while never reaching $x^2+y^2=1$). Or have I misunderstood something?






multivariable-calculus maxima-minima







share|cite|improve this question











$endgroup$




Consider the function



$$f(x,y) = e^xcos(y).$$



I am asked if this function has a maximum or a minimum inside the unit circle ${x,y: x^2+y^2<1}$.



The answer provided here is that, since the partial derivatives of the function are never $0$ simultaneously, the gradient is never zero, so the function has no minima or maxima. The answer is not in English so it would be difficult to quote it exactly here, but the wording implies there is no maxima or minima in any set.



But let's say we included the points ${x,y:x^2+y^2=1}$. Wouldn't the function now have a maxima at the "endpoints" $x^2+y^2=1$? To me it seems there is probably a miswording, and the author meant that the function has no maxima on the open set in question, as the function grows arbitrarily large as we approach the edge of the set (while never reaching $x^2+y^2=1$). Or have I misunderstood something?






multivariable-calculus maxima-minima




multivariable-calculus maxima-minima






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 28 at 21:22









J. W. Tanner

4,0271320




4,0271320










asked Jan 28 at 18:04









S. RotosS. Rotos

1644




1644








  • 1




    $begingroup$
    Note that the function is harmonic
    $endgroup$
    – Gabriele Cassese
    Jan 28 at 18:06










  • $begingroup$
    @gabrielecassese Did you see this just by looking at the function ? Or you went through the calculations ?
    $endgroup$
    – Thinking
    Jan 29 at 17:57










  • $begingroup$
    This kind of function ($e^xsin(y), e^ycos(x)$) is a quite often used example of harmonic function.
    $endgroup$
    – Gabriele Cassese
    Jan 29 at 18:09














  • 1




    $begingroup$
    Note that the function is harmonic
    $endgroup$
    – Gabriele Cassese
    Jan 28 at 18:06










  • $begingroup$
    @gabrielecassese Did you see this just by looking at the function ? Or you went through the calculations ?
    $endgroup$
    – Thinking
    Jan 29 at 17:57










  • $begingroup$
    This kind of function ($e^xsin(y), e^ycos(x)$) is a quite often used example of harmonic function.
    $endgroup$
    – Gabriele Cassese
    Jan 29 at 18:09








1




1




$begingroup$
Note that the function is harmonic
$endgroup$
– Gabriele Cassese
Jan 28 at 18:06




$begingroup$
Note that the function is harmonic
$endgroup$
– Gabriele Cassese
Jan 28 at 18:06












$begingroup$
@gabrielecassese Did you see this just by looking at the function ? Or you went through the calculations ?
$endgroup$
– Thinking
Jan 29 at 17:57




$begingroup$
@gabrielecassese Did you see this just by looking at the function ? Or you went through the calculations ?
$endgroup$
– Thinking
Jan 29 at 17:57












$begingroup$
This kind of function ($e^xsin(y), e^ycos(x)$) is a quite often used example of harmonic function.
$endgroup$
– Gabriele Cassese
Jan 29 at 18:09




$begingroup$
This kind of function ($e^xsin(y), e^ycos(x)$) is a quite often used example of harmonic function.
$endgroup$
– Gabriele Cassese
Jan 29 at 18:09










1 Answer
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$begingroup$

If you include the cincunference $x^2+y^2=1$, you are looking for minima/maxima of a continuous function over a compact set. Weierstrass's theorem guarantees their existence, and you can compute them using Lagrange multipliers.






share|cite|improve this answer









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    $begingroup$

    If you include the cincunference $x^2+y^2=1$, you are looking for minima/maxima of a continuous function over a compact set. Weierstrass's theorem guarantees their existence, and you can compute them using Lagrange multipliers.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      If you include the cincunference $x^2+y^2=1$, you are looking for minima/maxima of a continuous function over a compact set. Weierstrass's theorem guarantees their existence, and you can compute them using Lagrange multipliers.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        If you include the cincunference $x^2+y^2=1$, you are looking for minima/maxima of a continuous function over a compact set. Weierstrass's theorem guarantees their existence, and you can compute them using Lagrange multipliers.






        share|cite|improve this answer









        $endgroup$



        If you include the cincunference $x^2+y^2=1$, you are looking for minima/maxima of a continuous function over a compact set. Weierstrass's theorem guarantees their existence, and you can compute them using Lagrange multipliers.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 28 at 21:11









        PierreCarrePierreCarre

        1,695212




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