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Find local extrema of the function $$u(x,y,z)=\sin x \cdot \sin y\cdot \sin z$$ with the condition $$x+y+z=\frac{\pi}{2};\; x,y,z>0$$

Can anyone give me pointers on how to solve this problem?

Silvia Ghinassi
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zermelovac
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  • @MikePierce I formed the function $$F(x,y,z,\lambda)=\sin x ⋅\sin y⋅\sin z+\lambda \left(x+y+z-\frac{\pi}{2}\right) $$ and i found the derivatives of $F'x,; F'_y,; F'_z; \text{and}; F'\lambda$. Now I have to solve the system of equations: $$ F'x= F'_y=F'_z= F'\lambda=0$$. That's the part I really have no clue of how to solve. – zermelovac Sep 01 '15 at 13:50
  • The point is that you do not really need Lagrange's multipliers (notice that the domain is not closed, too), just basic considerations. – Jack D'Aurizio Sep 01 '15 at 13:54
  • Our TA has only explained this method of solving this kind of problem and on our exams we have to use Lagrange to solve the problem. – zermelovac Sep 01 '15 at 14:02
  • I will never understand the purpose of tying students' hands. Anyway, you may just eliminate $z$ and study a functions of two variables over a triangle. You have to study the boundary, since the domain is open, then to locate the stationary points inside by differentiation. – Jack D'Aurizio Sep 01 '15 at 14:05

2 Answers2

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$\sin(x)$ is positive on $\left(0,\frac{\pi}{2}\right)$, hence $\inf u=0$ and $u$ has no local minimum. On the other hand, by the AM-GM inequality and by the concavity of the sine function it follows that $x=y=z=\frac{\pi}{6}$ is the only local maximum.

Jack D'Aurizio
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1

Hint: By Jensen's inequality, $$\sum \log(\sin x)\le 3\log(\sin \frac{x+y+z}3)$$

Or if you must use the Lagrangian, you get from first order conditions: $$-\lambda = \cos x \sin y \sin z=\sin x \cos y \sin z = \sin x \sin y \cos z$$ $$\implies \tan x=\tan y =\tan z$$ $$\implies x = y=z\qquad\text{why?}$$ Hope you can fill in the rest...

Macavity
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