I try to find the vector $x = (x_1, \cdots, x_n)$ to maximize function $f(x)=f(x_1, \cdots, x_n)$ subject to the constraint $x_1^2 + \cdots +x_n^2 = a$, where $a$ is a positive constant. I use Lagrange multilplier and show that $df/dx_1=0$ at $x_1 = 0$. So can I conclude that the optimal solution $x$ must has the form $x = (0, x_2, \cdots, x_n)$?

  • $\begingroup$ The solutions of an optimization problem are usually NOT critical points of $f$. $\endgroup$ – Taladris May 28 '14 at 3:39

The FOC conditions you get are $df/dx_i-2\lambda x_i=0$. This does not mean that $\dfrac{df}{dx_i}=0$ at $x_i=0$.

Of course, if you are able to prove that $\frac{df}{dx_1}(0,x_2,\ldots,x_n)=0$ then you can test if there is a solution $x=(0,x_2,\ldots,x_n)$.

If $x_i\neq 0$ for all $i$ you can use $\dfrac{df/dx_j}{df/dx_i}=x_j/x_i$ and $x_1^2+\ldots+x_n^2=a$ and solve for the $x$s.


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