I have a hard time with Lagrange multiplier:
$$F(x,y)=x^2y$$
Constraint: $x+y=1$
Are there a max and min on these constraint?
Ackording to my calculations:
$$L(x,y,\lambda)= x^2y - \lambda(x+y-1)$$
I get:
$x=0$, $y= 0$ or $x=2/3$, $y=1/3$
According the solution: NO, there are no max/min.
Why does not the Lagrange Multiplier react to this? I am doing something wrong.
You can just plug in $y=1-x$ to the function to get $f(x,y)=x^2(1-x)=x^2-x^3$ along the curve. So the extremum occurs when $$\frac{d}{dx}(x^2-x^3)=2x-3x^2=x(2-3x)=0\implies x=0,2/3$$ as you said. The second derivative is $$\frac{d^2}{dx^2}(x^2-x^3)=2-6x$$ which is $2$ at $x=0$ and $-2$ at $x=2/3$. Therefore $x=0$ gives a local minimum and $x=2/3$ gives a local maximum.
However, there are no global extrema because the function's range is all of $\mathbb{R}$. It is an odd polynomial.