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I have to find maxima and minima of: $f(x,y)=\cos(x)+\cos(y)$, bounded by $\Bbb D: x^{4}+y^{4}=1$.

  1. $f(x,y)$ is "enough" smooth, and $\Bbb D$ is a compact set $\Rightarrow$ $\text{(max, min)}$ must exist.
  2. There is no internal part for $\Bbb D$; so I don't have to find $(x,y)\ | \ \nabla f(x,y)= (0,0)$.
  3. If $g(x,y)=x^{4}+y^{4}-1$, there are no points which satisfy $\begin{cases} \nabla g(x,y)=(0,0) \\ g(x,y)=0 \end{cases}$.
  4. Applying Lagrange multipliers method, I remain stuck into $\begin{cases} -\sin(x)=4\lambda x^{3} \\ -\sin(y)=4\lambda y^{3} \\ x^{4}+y^{4}=1 \end{cases}$.

I wasn't able to solve that system $\text{wrt} (x,y)$.
Is there another way to find (max, min) avoiding that system?
Or maybe some useful trick to writing that in a simpler form?

Thanks.

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    $\begingroup$ Are you looking for the min and max of $f$ on the curse $x^4+y^4 = 1$ or in the area bounded by this curve (so $x^4+y^4 ≤ 1$)? $\endgroup$
    – LL 3.14
    Apr 30 '20 at 18:42
  • $\begingroup$ Sorry if I can't explain well.. I'm searching on $x^{4} + y^{4} = 1$. $\endgroup$ Apr 30 '20 at 19:48
  • $\begingroup$ You can only approximate them. $\endgroup$
    – Allawonder
    May 1 '20 at 11:47
  • $\begingroup$ Anyway, one of the local extrema is easy to determine, namely $1+\cos 1.$ $\endgroup$
    – Allawonder
    May 1 '20 at 11:47
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The equations $-\sin x=4\lambda x^3$, $\>-\sin y=4\lambda y^3$ are solved by the admissible points $(\pm1,0)$ and $(0,\pm1)$, whereby $\lambda=-{1\over4}\sin 1$ in each case. But there may be other solutions where both $x$ and $y$ are nonzero. In such a case we have $$x^2\cdot{x\over\sin x}=y^2\cdot{y\over\sin y}\tag{1}$$ for such a point. Drawing the plot of the function $$t\mapsto t^2\cdot{t\over\sin t}\qquad(-1\leq t\leq1)$$ we see that it looks like a parabola. In particular $(1)$ implies $y=\pm x$. Together with $x^4+y^4=1$ we therefore obtain four more conditionally critical points, namely $\bigl(\pm 2^{-1/4}, \pm 2^{-1/4}\bigr)$.

enter image description here

At the first four points we have $f(\pm1,0)=f(0,\pm1)=1+\cos1=1.5403$, and at the other four points we have $f\bigl(\pm 2^{-1/4}, \pm 2^{-1/4}\bigr)=2\cos 2^{-1/4}=1.3336$. It follows that at the first four points $f$ assumes its maximum, and at the second four points its minimum on the admissible set.

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  • $\begingroup$ So, by the fact: "it looks like a parabola", I'm sure that for every $y \in [0,1]$, there are only 2 values of $x$ which confirm the equality, namely $+x$ and $-x$.. Right? $\endgroup$ May 1 '20 at 14:15
  • $\begingroup$ @DOmonoXYLEDyL: Yes. See my edit. $\endgroup$ May 1 '20 at 15:05
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Clearly the maximum occurs at $(0,0)$:

enter image description here

as simple derivation will prove.

The minimum occurs where $x = \pm y$ and on the boundary $x^4 + y^4 = 1$, which again is a simple matter of derivatives: $2 \cos^4 x = 1$ or $\cos x = \sqrt[4]{1/2}$ or $x = \pm y = \cos^{-1} (\sqrt[4]{1/2})$.

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  • $\begingroup$ Sorry, maybe I didn't explain much well... I'm searching on $x^{4}+y^{4}=1$, so $(0,0)$ can't be the maximum. And also, why are you stating that minimum is where $x= \pm y$? (intersected with $x^{4}+y^{4}=1$, I guess). I mean, you're clearly right (I've graphed the system, and solutions are the same)… but how did you derive just $x= \pm y$? $\endgroup$ Apr 30 '20 at 19:58
  • $\begingroup$ Ok, I understood: from $\frac{\sin{x}}{\sin{y}} = \frac{x^{3}}{y^{3}}$, $x=0 \Rightarrow 0=0 \; \forall y \in \Bbb R$, $x=y \Rightarrow 1=1$, $x=-y \Rightarrow -1 = -1$; so, each one is a solution. Then, how the uniqueness of these solutions is guaranteed? $\endgroup$ Apr 30 '20 at 21:48
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    $\begingroup$ Oh... then don't write bounded by but instead constrained to obey...!!! $\endgroup$ Apr 30 '20 at 22:36
  • $\begingroup$ I'm really Sorry, but I don't speak English and I do my best to be much clear as possibile when I write my questione.. But evidently, it's not enough.. $\endgroup$ May 1 '20 at 8:04

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