Is it the absolute maximum?

Question

Let $x,y$ and $z$ be the interior angles of a triangle. Suppose that $f(x,y,z)=\sin(x)\sin(y)\sin(z)$. Using the Lagrange multipliers, I know $f$ has a local maximum at $x=y=z=\pi/3$. However, I do not know if it is the global maximum.

Answer 1

Since the gradient of the constraint $g(x,y,z)=x+y+z-\pi$ $$\bigtriangledown g=<1,1,1>\neq 0$$ The method of Lagrange Multiplier ensures that if $f$ has a global maximum then it must occur at the point you found. As gpassante pointed out, $f$ has a global maximum on $K$, the set of all $(x,y,z)$ that satisfies the constraint. But in general, the point you found via Lagrange Multiplier might very well be a minimum (consider for example $f'(x,y,z)=-\sin(x)\sin(y)\sin(z)$). Luckily we can rule out this possibility by simply picking another point in $K$ and showing that $f$ is smaller at that point.

Answer 2

This isn't a proof, but you can just plot the function and see that it is indeed the local maximum. Notice that $f(x,y,z) = f(x,y,\pi-x-y) = \sin(x)\sin(y)sin(x+y)$. If you plot this function you will see that $x=y=z=\pi/3$ is the maximum.

https://www.wolframalpha.com/input/?i=sin%28x%29+sin%28y%29+sin%28x+%2B+y%29%2C+from+x%3D+0+to+pi%2C+y%3D0+to+pi