Extrema of function - What is $\lambda$?

Question

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 $x_1, x_2, x_3>0$.

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*}

So we get the following: \begin{equation*} \tilde{f}_{x_1x_1}(x_1, m-4x_1)=-36x_1\cdot (m-4x_1), \tilde{f}_{x_2x_2}(x_1,m-4x_1)=-18x_1, \tilde{f}_{x_1x_2}(x_1,m-4x_1)=9m -36x_1-18 (m-4x_1)\end{equation*} So, it holds that \begin{equation*}\tilde{f}_{x_1x_1}(x_1, m-4x_1)\tilde{f}_{x_2x_2}(x_1, m-4x_1) - \tilde{f}^2_{x_1x_2}(x_1, m-4x_1)=18\cdot 36x_1^2(m-4x_1)-(9m -36x_1-18 (m-4x_1))^2\end{equation*} For some specific $m$ it is $>0$, then the function has extrema, right?

$$$$

Then we have to calculate $x_1^{\star}(m), x_2^{\star}(m), \lambda^{\star}(m)$.

What is $\lambda$ ?

Answer 1

The fact that your problem mentions $\lambda$ says to me that Lagrange multipliers are the preferred solution method. Let $g(x_1,x_2,x_3) = 2x_1 + x_2 + x_3$. You are asked to find the maximum of $f$ on the set constrained by $g=m$, $x_1>0$, $x_2>0$, $x_3>0$. The Lagrange multiplier equations are $\nabla f = \lambda \nabla g$, or \begin{align} 9x_2 x_3 &= 2\lambda \tag{1}\label{1} \\ 9x_1 x_3 &= \lambda \tag{2}\label{2} \\ 9x_1 x_2 &= \lambda \tag{3}\label{3} \\ \end{align} along with the constraint equation $$ 2x_1 + x_2 + x_3 = m \tag{4}\label{4} $$ Comparing the left-hand sides of \eqref{2} and \eqref{3}, we get $9x_1x_3 = 9x_1 x_2$. Since $x_1 > 0$, we can cancel it and get $x_2 = x_3$. Similarly, comparing the left-hand sides of \eqref{1} and \eqref{2}, we get $9x_2 x_3 = 2(9x_1 x_3)$, so $x_2 = 2x_1$. Hence $x_3 = 2x_1$ as well. Substituting these into \eqref{4} gives $$ 2x_2 + 2x_1 + 2x_1 = m \implies x_1 = \frac{m}{6} $$ and therefore $x_2 = x_3 = \frac{m}{3}$. In the notation of your problem: \begin{align*} x_1^\star(m) &= \frac{m}{6} \\ x_2^\star(m) &= \frac{m}{3} \\ x_3^\star(m) &= \frac{m}{3} \\ \lambda^\star(m) &= 9 x_1^\star(m) x_2^\star(m) = 9\cdot \frac{m}{6} \cdot\frac{m}{3} = \frac{m^2}{2} \end{align*}

You asked:

What is $\lambda$?

and I think you meant, what is $\lambda$ in terms of $m$ in this problem? You have your answer to that. But the deeper question is whether this seemingly arbitrary variable inserted into the solution method has any meaning. And it does.

Let $f^\star(m)$ be the constrained maximum value of $f$, as a function of $m$. We know $$ f^\star(m) = f(x_1^\star(m),x_2^\star(m),x_3^\star(m)) = 9 \cdot \frac{m}{6}\cdot \frac{m}{3} \cdot \frac{m}{3} = \frac{m^3}{6} $$ Notice that $$ \frac{df^\star}{dm} = \frac{m^2}{2} = \lambda $$ This is true in general. In words, the Lagrange multiplier is the change in the constrained maximum as the constraint level changes. In economics, a lot of times $f$ is a utility function and $g$ is a cost function. Then $\lambda$ is the marginal utility of money.

Answer 2

Firstly, observe that the function has no minimum. It can go as close to $0$ as you want but never reaches $0$.

Secondly, see that $x_2$ and $x_3$ are symmetric. So the AMGM inequality says, fixing any $x_1$, the maximum is given by $x_2=x_3$.

Therefore, you are left with the following problem: maximize $9x_1 \times y^2$ under the constraint $2x_1+2 y=m$ (for we have set $x_2=x_3=y$). At this stage, the problem can be solved with single-variable calculus by substituting the constraint to the objective function.