I want to solve the constrained minimization problem
$$
\begin{align}
\min_x \quad &\cos(x) \\
\text{s.t.} \quad & 0 \leq x \leq 2 \pi
\end{align}
$$
with Lagrange multipliers. Of course, the minimum is $-1$ at $x = \pi$, but I don't get that solution. My attempt:
$$
\begin{align}
\min_x \quad &f(x) \\
\text{s.t.} \quad & g_1(x) \leq 0 \\
& g_2(x) \leq 0
\end{align}
$$
can be solved by taking the Lagrange function:
$$
L(x, \lambda_1, \lambda_2) = f(x) + \lambda_1 g_1(x) + \lambda_2 g_2(x)
$$
and setting its gradient to zero. In my example I have:
$$
\begin{align}
f(x) &= \cos(x) \\
g_1(x) &= -x \\
g_2(x) &= x – 2 \pi
\end{align}
$$
So the partial derivatives give me these equations:
$$
\begin{align}
\frac{\partial}{\partial x} L(x, \lambda_1, \lambda_2) &= -\sin(x) – \lambda_1 + \lambda_2 = 0 \\
\frac{\partial}{\partial \lambda_1} L(x, \lambda_1, \lambda_2) &= -x = 0 \\
\frac{\partial}{\partial \lambda_2} L(x, \lambda_1, \lambda_2) &= x – 2 \pi = 0
\end{align}
$$
But this is bad, because from the second equation I get that $x = 0$ is a solution and from the third that $x = 2 \pi$ is a solution. But $0 \neq 2 \pi$ so there is something wrong because I should get $x = \pi$.
Where is the mistake here?
Best Answer
As already stated in the comments, the method of Lagrange multipliers serves to incorporate equality constraints into an optimization problem. You can use it to solve optimization problems with inequality constraints by finding the optima in the interior of the feasible region, finding the optima on the boundary (using one or more inequality constraints as equality constraints in the method of Lagrange multipliers) and comparing the function values to find the global optimum.
In the present case, the optimum in the interior is $-1$ at $x=0$. The boundaries are single points, so there's nothing to optimize there; nevertheless, if you want you can apply the method of Lagrange multipliers roughly as you did (but separately in two separate applications of the method, once for each boundary) to recover the boundary points at $x=0$ and $x=2\pi$. Comparing the function values shows that the global optimum is the one in the interior.