0
$\begingroup$

somehow I'm not able to solve this:

$f:\mathbb R^2 \to \mathbb R, \quad (x,y)\mapsto 2x^3-12x+3y^2+6xy$

$D:=\{(x,y)\in\mathbb R^2 | x\geq 0, y\geq 0 , x+y\leq 1\}$

Get the extrema on $f|_D$

Solution:

Define $g(x,y):=x+y-1=0$

Define $L=f-\lambda g=2x^3-12x+3y^2+6xy-\lambda(x+y-1)$

$\frac{\partial L}{\partial x} = 6x^2-12+6y-\lambda \overset{!}{=} 0$

$\frac{\partial L}{\partial y} =6y+6x-\lambda \overset{!}{=} 0$

We get

$\lambda = 6x^2-12+6y$ $\lambda=6y+6x$ $g=0$

So

$\lambda/6=x^2-2+y$ $\lambda/6=y+x$ $1=x+y$

It follows: $\lambda = 6$ und $y=1-x$

We get: $1=x^2-2+y=x^2-2+1-x \Rightarrow 0=x^2-2-x=(x-2)(x+1)$

because $x\geq 0 \Rightarrow x_1=2 \Rightarrow y=-1$

But because y must be greater zero, we don't find any point.

So whats wrong here?

  • 0
    $D$ is not defined by $x + y - 1 = 0$. $D$ is a triangular region. This is not the kind of problem for which Lagrange multipliers is appropriate. Just find the critical points of $f$, if any, in the interior of $D$. And also check for extrema of $f$ on the boundary (by substitution), one piece at a time. – 2017-01-18
  • 0
    Yeah, I also though about that it's inappropreate here but I still struggle a bit knowing when to use the lagrangian multipliers. But good to know that was the error here, thanks! – 2017-01-18
  • 0
    @quasi Why aren't Lagrange multipliers appropriate? Can't they deal with inequality constraints via the KKT conditions? – 2017-01-18
  • 0
    @LinAlg -- Maybe, but there's no need. What possible gain in simplicity could it yield? The domain $D$ is a very simple triangular region in $\mathbb{R}^2$. The critical points in the interior of $D$ can be found in the usual way by setting the partial derivatives of $f$ to $0$ and solving for $x,y$. The boundary needs 3 cases, but in each case, the optimization of $f$ reduces to an easy 1-variable optimization problem. – 2017-01-18

0 Answers 0