Extremum of $f:(\mathbb{R_+})^n \to \mathbb{R}$, $f(x_1,\dots ,x_2)=(1+x_1)\cdots(1+x_n)$ with $x_1 \cdots x_n=a^n$ for a fixed $a>0$

Question

$f:(\mathbb{R_+})^n \to \mathbb{R}$

$f(x_1,\dots ,x_2)=(1+x_1)\cdots(1+x_n)$

I want to determine where $f$ has a local or global extremum with the condition $x_1 \cdots x_n=a^n$ for a fixed $a>0$ using lagrange multipliers. How can this be done?

Answer 1

The general strategy would be as follows:

• Compute the gradient of $f$, i.e. $\nabla f$
• Give a name to your constraint, e.g. $g$ and compute its gradient too, i.e. $\nabla g$
• Introduce a Lagrange multiplier $\lambda$ and consider the two equations $$ \begin{cases} \nabla f + \lambda \nabla g & = 0 \\ g & = 0 \end{cases} $$ to be solved on the domain of $f$, i.e. $(\mathbb R_+)^n$.

That being said, if seems that you would have an easier time working with logarithms for that particular problem, which may also give you an intuition about the solutions.

Answer 2

Calling

$$ L(x,\lambda) = \prod_{k=1}^n(1+x_k)-\lambda\left(\prod_{k=1}^n x_k - a^n\right) $$

we have the stationary points as solutions for

$$ \nabla L =\begin{cases} \frac{\prod_{k=1}^n(1+x_k)}{1+x_k}-\frac{\lambda \prod_{k=1}^n x_k}{x_k} = 0,\ \ k = 1,\cdots ,n\\ \prod_{k=1}^n x_k - a^n = 0 \end{cases} $$

now calling

$$ \mu = \lambda\frac{\prod_{k=1}^n x_k}{\prod_{k=1}^n(1+x_k)} $$

we have

$$ \frac{1}{1+x_k}-\frac{\mu}{x_k} = 0,\ \ k = 1,\cdots, n $$

or

$$ x_k = \frac{1-\mu}{\mu}\Rightarrow \frac{1-\mu}{\mu} = a \Rightarrow \mu = \frac{1}{a+1} $$

hence the solution $x_k = a$ is extremal