- #1
mathmari
Gold Member
MHB
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Hey! 
We have the function $f(x_1, x_2, x_3)=9x_1\cdot x_2\cdot x_3$ and we want to find possible extremas under the constraint $2x_1+x_2+x_3=m, m>0$ and $x_1, x_2, x_3>0$.
Then I have to calculate $x_1^{\star}(m), x_2^{\star}(m), \lambda^{\star}(m)$. I have done the following:
\begin{equation*}2x_1+x_2+x_3=m \Rightarrow x_3=m-2x_1-x_2\end{equation*}
\begin{equation*}\tilde{f}(x_1, x_2)=9x_1\cdot x_2\cdot (m-2x_1-x_2)=9mx_1\cdot x_2-18x_1^2\cdot x_2-9x_1\cdot x_2^2\end{equation*}
\begin{align*}\tilde{f}_{x_1}=9m x_2-36x_1\cdot x_2-9 x_2^2 \\ \tilde{f}_{x_1x_1}=-36x_1\cdot x_2 \\ \tilde{f}_{x_2}=9mx_1-18x_1^2-18x_1\cdot x_2 \\ \tilde{f}_{x_2x_2}=-18x_1 \\ \tilde{f}_{x_1x_2}=9m -36x_1-18 x_2\end{align*}
\begin{equation*}\tilde{f}_{x_1}=0 \Rightarrow 9m x_2-36x_1\cdot x_2-9 x_2^2=0\Rightarrow x_2=0 \text{ or } x_2=m-4x_1\end{equation*}
\begin{equation*}\tilde{f}_{x_1}=0 \Rightarrow 9m x_2-36x_1\cdot x_2-9 x_2^2=0\Rightarrow x_2=0 \text{ or } x_2=m-4x\end{equation*}
$x_2$ does not satisfy the constraint, so it is rejected, and therefore it must hold $x_2=m-4x_1$. So, then $x_3=m-2x_1-(m-4x_1)=m-2x_1-m+4x_1=2x_1$.
So, we get the function \begin{equation*}g(x_1)=f(x_1, m-4x_1, 2x_1)=9x_1\cdot (m-4x_1)\cdot 2x_1=18x_1^2\cdot (m-4x_1)\end{equation*}
The first derivative is \begin{equation*}g'(x_1)=36x_1\cdot (m-4x_1)+18x_1^2\cdot (-4)=36mx_1-144x_1^2-72x_1^2=36mx_1-216x_1^2\end{equation*}
and the roots are $x_1=0$, that doesn't satisfy the constraint, and $x_1=\frac{m}{6}$. So, the extremum is at the point \begin{align*}(x_1^{\star}(m), x_2^{\star}(m), x_3^{\star}(m)) & =\left (\frac{m}{6}, m-4\frac{m}{6}, 2\frac{m}{6}\right )=\left (\frac{m}{6}, m-\frac{2}{3}m, \frac{1}{3}m\right )=\left (\frac{m}{6}, \frac{m}{3}, \frac{m}{3}\right )\end{align*}
and it is equal to \begin{equation*}f^{\star}(m)=9\cdot \frac{m}{6}\cdot \frac{m}{3}\cdot \frac{m}{3}=\frac{m^3}{6}\end{equation*}
We have that $\tilde{f}_{x_1x_1}\left (\frac{m}{6}, \frac{m}{3}\right )=-36\frac{m}{6}\frac{m}{3}=-2<0$.
So $\tilde{f}$ has a maximum at $\left (\frac{m}{6}, \frac{m}{3}\right )$.
Therefore $f$ has a maximum at $\left (\frac{m}{6}, \frac{m}{3}, \frac{m}{3}\right )$. is eveything correct? (Wondering) What is $\lambda$ ?
I have to show also that $\frac{df^{\star}}{dm}=\lambda (m)$. (Wondering)
We have the function $f(x_1, x_2, x_3)=9x_1\cdot x_2\cdot x_3$ and we want to find possible extremas under the constraint $2x_1+x_2+x_3=m, m>0$ and $x_1, x_2, x_3>0$.
Then I have to calculate $x_1^{\star}(m), x_2^{\star}(m), \lambda^{\star}(m)$. I have done the following:
\begin{equation*}2x_1+x_2+x_3=m \Rightarrow x_3=m-2x_1-x_2\end{equation*}
\begin{equation*}\tilde{f}(x_1, x_2)=9x_1\cdot x_2\cdot (m-2x_1-x_2)=9mx_1\cdot x_2-18x_1^2\cdot x_2-9x_1\cdot x_2^2\end{equation*}
\begin{align*}\tilde{f}_{x_1}=9m x_2-36x_1\cdot x_2-9 x_2^2 \\ \tilde{f}_{x_1x_1}=-36x_1\cdot x_2 \\ \tilde{f}_{x_2}=9mx_1-18x_1^2-18x_1\cdot x_2 \\ \tilde{f}_{x_2x_2}=-18x_1 \\ \tilde{f}_{x_1x_2}=9m -36x_1-18 x_2\end{align*}
\begin{equation*}\tilde{f}_{x_1}=0 \Rightarrow 9m x_2-36x_1\cdot x_2-9 x_2^2=0\Rightarrow x_2=0 \text{ or } x_2=m-4x_1\end{equation*}
\begin{equation*}\tilde{f}_{x_1}=0 \Rightarrow 9m x_2-36x_1\cdot x_2-9 x_2^2=0\Rightarrow x_2=0 \text{ or } x_2=m-4x\end{equation*}
$x_2$ does not satisfy the constraint, so it is rejected, and therefore it must hold $x_2=m-4x_1$. So, then $x_3=m-2x_1-(m-4x_1)=m-2x_1-m+4x_1=2x_1$.
So, we get the function \begin{equation*}g(x_1)=f(x_1, m-4x_1, 2x_1)=9x_1\cdot (m-4x_1)\cdot 2x_1=18x_1^2\cdot (m-4x_1)\end{equation*}
The first derivative is \begin{equation*}g'(x_1)=36x_1\cdot (m-4x_1)+18x_1^2\cdot (-4)=36mx_1-144x_1^2-72x_1^2=36mx_1-216x_1^2\end{equation*}
and the roots are $x_1=0$, that doesn't satisfy the constraint, and $x_1=\frac{m}{6}$. So, the extremum is at the point \begin{align*}(x_1^{\star}(m), x_2^{\star}(m), x_3^{\star}(m)) & =\left (\frac{m}{6}, m-4\frac{m}{6}, 2\frac{m}{6}\right )=\left (\frac{m}{6}, m-\frac{2}{3}m, \frac{1}{3}m\right )=\left (\frac{m}{6}, \frac{m}{3}, \frac{m}{3}\right )\end{align*}
and it is equal to \begin{equation*}f^{\star}(m)=9\cdot \frac{m}{6}\cdot \frac{m}{3}\cdot \frac{m}{3}=\frac{m^3}{6}\end{equation*}
We have that $\tilde{f}_{x_1x_1}\left (\frac{m}{6}, \frac{m}{3}\right )=-36\frac{m}{6}\frac{m}{3}=-2<0$.
So $\tilde{f}$ has a maximum at $\left (\frac{m}{6}, \frac{m}{3}\right )$.
Therefore $f$ has a maximum at $\left (\frac{m}{6}, \frac{m}{3}, \frac{m}{3}\right )$. is eveything correct? (Wondering) What is $\lambda$ ?
I have to show also that $\frac{df^{\star}}{dm}=\lambda (m)$. (Wondering)
