Extrema of function - What is λ?

[mathmari]

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)

 

[mathmari]

Ah we have to use Lagrange multipliers, right? (Wondering)

 

[HallsofIvy]

I wondered where "\lambda" came from! There is no "\lambda" in the statement of the problem so if the problem asks you to find it then, yes, it is referring to the Lagrange multiplier method.