MHB Are the critical points minima or maxima?

[mathmari]

Hey!

We have the function $f(x_1, x_2)=2-x_1-x_2$ and we want to check if it has maxima or minima under the constraint $x_1^2+x_2^2=8$. Since we cannot solve for one variable at the equationof the constraint, we have to use the Langrange function, right? (Wondering)

I have done the following:

Let $g(x_1, x_2)=x_1^2+x_2^2-8$.

• \begin{equation*}L(x_1,x_2,\lambda )=2-x_1-x_2 -\lambda \cdot \left (x_1^2+x_2^2-8\right )\end{equation*}
• \begin{align*}&L_{x_1}(x_1,x_2,\lambda)=-1 -2x_1\lambda \\ & L_{x_2}(x_1,x_2,\lambda)=-1 -2x_2\lambda \\ & L_{\lambda}(x_1,x_2,\lambda)=- \left (x_1^2+x_2^2-8\right )\end{align*}
• \begin{align*}&L_{x_1}(x_1,x_2,\lambda)=0 \Rightarrow -1 -2x_1\lambda=0 \\ & L_{x_2}(x_1,x_2,\lambda)=0 \Rightarrow -1 -2x_2\lambda=0 \\ & L_{\lambda}(x_1,x_2,\lambda)=0 \Rightarrow =- \left (x_1^2+x_2^2-8\right )=0\end{align*}
• From the first equation we have that \begin{equation*}2x_1\lambda=-1 \Rightarrow \lambda=-\frac{1}{2x_1} \end{equation*}
  
  Replacing this in the second equation we get \begin{equation*}-1 -2x_2\left (-\frac{1}{2x_1}\right )=0 \Rightarrow -1 +\frac{x_2}{x_1}=0 \Rightarrow \frac{x_2}{x_1}=1 \Rightarrow x_2=x_1 \end{equation*}
  
  Since $x_2=x_1$ from the third equation we get \begin{equation*}x_1^2+x_1^2-8=0 \Rightarrow 2x_1^2=8 \Rightarrow x_1^2=4 \Rightarrow x_1=\pm 2\end{equation*}
  
  Therefore we get that's $({x_1}_0, {x_2}_0)=(-2,-2)$ und $({x_1}_0, {x_2}_0)=(2,2)$.
  
  For ${x_1}_0=-2$ we get $\lambda=\frac{1}{x_1}=-\frac{1}{2}$ and for ${x_1}_0=2$ we get $\lambda=\frac{1}{x_1}=\frac{1}{2}$. So, the critical points are $(-2,-2)$ und $(2,2)$.

Since the partial derivatives of second order are zero, we cannot use the condition
\begin{equation*}f_{x_1x_1}(x_0, y_0)\left\{\begin{matrix}
<0\\
>0
\end{matrix}\right. \ \text{ and } \ f_{x_1x_1}(x_0, y_0)f_{x_2x_2}(x_0, y_0)-\left (f_{x_1x_2}(x_0, y_0)\right )^2>0\end{equation*} to check if we have maxima or minima.

What can we do then in tis case? (Wondering)

 

[MarkFL]

What I would do is observe that we have cyclical symmetry, that is we may interchange $x_1$ and $x_2$ with no change in either the objective function or the constraint, so we know the critical points are:

$$(-2,-2),\,(2,2)$$

Since:

$$f(-2,-2)=6$$ and $$f(2,2)=-2$$

we may therefore conclude:

$$f_{\min}=f(2,2)=-2$$

$$f_{\max}=f(-2,-2)=6$$

 

[mathmari]

What I would do is observe that we have cyclical symmetry, that is we may interchange $x_1$ and $x_2$ with no change in either the objective function or the constraint, so we know the critical points are:

$$(-2,-2),\,(2,2)$$

Since:

$$f(-2,-2)=6$$ and $$f(2,2)=-2$$

we may therefore conclude:

$$f_{\min}=f(2,2)=-2$$

$$f_{\max}=f(-2,-2)=6$$

Ah ok. So, it cannot be that we have a saddle point? (Wondering)

 

[MarkFL]

Ah ok. So, it cannot be that we have a saddle point? (Wondering)

The objective function is linear, so it describes a plane, and so I don't believe there will be any saddle points, or in fact any critical points without some kind of constraint imposed. :D

 

[mathmari]

The objective function is linear, so it describes a plane, and so I don't believe there will be any saddle points, or in fact any critical points without some kind of constraint imposed. :D

I see! Thank you very much! (Happy)