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Find the critical points of $ x + y^2 $ with the restriction $ 2x^2 + y^2 = 1 $ Use the bordered hessian matrix to classify the critical points.

So, using that $ F(x,y,\lambda) = f(x,y) - \lambda(g(x,y)) $ we have that

$ F(x,y, \lambda) = x + y^2 - 2\lambda x^2 - \lambda y^2 + \lambda $

And the bordered Hessian Matrix would be

$$ \begin{pmatrix} 0 & 4x & 2y \\ 4x & -4\lambda & 0 \\ 2y & 0 & 2 - 2\lambda \\ \end{pmatrix} $$

Next, the determinant would be

$ 4y^2(-4\lambda) - 16x^2(2-2\lambda) $

If the first derivatives equal to 0, we have that

$ 1 - 4\lambda x = 0 $

$ 2y - 2\lambda y = 0 \Rightarrow 2y (1 - \lambda ) = 0 \Rightarrow y = 0 $ or $ \lambda = 1$

If $\lambda = 1$, then $ x = {1 \over 4} $

From this, I get a critical point, I think, which is $ ({1 \over 4}, 0) $ . Evaluating the determinant in this point gives

$ 4y^2(-4\lambda) - 16x^2(2-2\lambda) = 2\lambda - 2$

Next I think that I have to analyse $\lambda$ (if it is minor than cero or equal than cero, to see if I obtain a minimum an maximun value? It is also not clear to me when I get a chair point or a point that I cannot decide what it is.

Trux
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1 Answers1

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Your derivatives w.r.t x and y are o.k. Also your Bordered Hessian Matrix. But you haven´t found all critical points.

You have also to calculate the partial derivative w.r.t. $\lambda$.

$\frac{\partial F}{\partial \lambda}=1-2x^2-y^2=0 \quad (1)$

The two other derivatives:

$\frac{\partial F}{\partial x}= 1 - 4\lambda x = 0 \quad (2)$

$\frac{\partial F}{\partial y}= 2y - 2\lambda y = 0 \quad (3)$

Case 1: $y=0$

Now we can insert $y=0$ in $(1)$: $1=2x^2 \Rightarrow x_{1/2}=\pm \frac{1}{\sqrt{2}}$ Thus the first two crititcal points are $P_1(\frac{1}{\sqrt{2}},0)$ and $P_2(-\frac{1}{\sqrt{2}},0)$

Case 2: $y\neq 0$

From (3) we get $\lambda=1$, because we can divide (3) by $2y$ and get $1-\lambda=0$. If $\lambda=1$, then we get from (2) that $x=\frac1 4$

If you insert the value for x in (1) you will get $y_{3/4}=\pm \sqrt{\frac7 8}$

Thus the next two crititcal points are $P_3(\frac1 4,\sqrt{\frac7 8})$ and $P_4(\frac1 4,-\sqrt{\frac7 8})$

The decision rule for a critical point, $x_0$, by using the Bordered Hessian Matrix, $\tilde{H}$, is:

  • local maximum, if $\text{det} \ \tilde{H}(x_0)>0$

  • local minimum, if $\text{det} \ \tilde{H}(x_0)<0$

  • no decision possible, if $\text{det} \ \tilde{H}(x_0)=0$

callculus42
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