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

Question

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.

Answer 1

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.

Answer 2

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