# Finding maxima and minima of $\cos(x)+\cos(y)$ constrained to obey $x^{4}+y^{4}=1$

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.

• 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 $x^4+y^4 ≤ 1$)? Apr 30 '20 at 18:42
• Sorry if I can't explain well.. I'm searching on $x^{4} + y^{4} = 1$. Apr 30 '20 at 19:48
• You can only approximate them. May 1 '20 at 11:47
• Anyway, one of the local extrema is easy to determine, namely $1+\cos 1.$ May 1 '20 at 11:47

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 $$\bigl(\pm 2^{-1/4}, \pm 2^{-1/4}\bigr)$$.

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.

• 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? May 1 '20 at 14:15
• @DOmonoXYLEDyL: Yes. See my edit. May 1 '20 at 15:05

Clearly the maximum occurs at $$(0,0)$$:

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 $$x = \pm y = \cos^{-1} (\sqrt[4]{1/2})$$.

• 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 $x= \pm y$? Apr 30 '20 at 19:58
• 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? Apr 30 '20 at 21:48
• Oh... then don't write bounded by but instead constrained to obey...!!! Apr 30 '20 at 22:36
• 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.. May 1 '20 at 8:04