- #1
mathmari
Gold Member
MHB
- 5,049
- 7
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*}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)
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 $(2,2)$.
\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)