Lagrangian Multipliers Constrained Optimization As far as I understand, Lagrangian multiplier $\lambda$ can take negative and positive values. For the positive values, we find maximum point. For the negative values, we find minimum point. 
I think that we cannot find minimum point by using Lagrangian multipliers for this functions:
$f(x, y) = x^2y^2$
$g(x, y) = x^2 + y^2 = 1$
I aim to find minimum and maximum point of $f$ under the condition $g$.
When I apply Lagrangian method, I could find only this solution:
$x = \sqrt\lambda$
$y = \sqrt\lambda$
When I substituted these equalities in $g(x, y)$, I found $\lambda = 1/2$, and $f(x, y) = 1/4$. 
What do you think about this ?
There is only positive Lagrangian value, and it makes function $f(x, y)$ produce $1/4$. Is this maximum value or minimum value ? I think it is maximum point. 
 A: Step by step
$$
L = x^2y^2-\lambda(x^2+y^2-1)
$$
so the stationary points (maxima, minima, saddle) are included in the solutions for
$$
\nabla L = 0 =\cases{2xy^2-2\lambda x\\
2y x^2-2\lambda y\\
x^2+y^2-1}
$$
from the first two equations we have $x=0,y=0,y^2=\lambda, x^2 = \lambda$ so substituting into the third we have $2\lambda = 1$ or $\lambda^* = \frac 12$ hence $x^* = \pm\frac{\sqrt 2}{2},\ \ y^* = \pm\frac{\sqrt 2}{2}$ or for any $\lambda$ also $x^* = 0,y^* =\pm 1,\ \ y^*=0,x^*= \pm 1$. Now among those stationary points we should choose according to our needs. Substituting the restriction on the objective function we have
$$
z = x^2(1-x^2),\ \ -1\le x \le 1
$$
so at $x^*=\pm\frac{\sqrt 2}{2}$ we have two maxima and at $x^* = 0, x^*=\pm 1$ we have minima as can be depicted from the following plot

Resuming the stationary points
$$
\begin{array}{ccccc}
x^* & y^* & \lambda^* & x^2y^2 & \text{type}\\
0  & \pm 1 & \text{any} & 0 & \text{min}\\
\pm 1  & 0 & \text{any} & 0 & \text{min}\\
\pm\frac{\sqrt 2}{2} & \pm\frac{\sqrt 2}{2} & \frac 12 & \frac 14 & \text{max}
\end{array}
$$
NOTE
The restriction of $z = x^2y^2$ on the cylinder $x^2+y^2-1=0$ can be depicted in blue in the following plot.

A: You have found the maximum. 
When solving the Euler-Lagrange equations, you ignored that the derivatives of the Lagrangian with respect to $x$ and $y$ can also be zero when $(x,y)=(0,\sqrt\lambda)$ or $(x,y)=(\sqrt\lambda,0)$. Then you get $\lambda=1$, and $$f(x,y)=0$$which is the minimum.

