Using Lagrange multipliers, find the maximum value of a square root

Question: Let $f(x_1, x_2,\ldots, x_n)=\sqrt[n]{x_1x_2\cdots x_n}$, where $x_1, \ldots, x_n$ are positive numbers and $x_1+\cdots+x_n=c$, where $c$ is a constant. Using Laplace multipliers, find the maximum of $f$.

Let $g(x_1, x_2,\ldots, x_n)=x_1+\cdots+x_n$. By Laplace's theorem, we can find λ such that $∇f=λ∇g$. However, since $x_1, \ldots, x_n$ are numbers, then $∇g=∇f=0$. How am I supposed to find the maximum? (which is $\frac{c}{n}$ according to my textbook's answer key).

• The $x_i$ are viewed as variables, not fixed numbers, to do this. – coffeemath Oct 17 '15 at 0:56

I will point out the steps you should follow to do this problem because it seems that you may be new to Lagrange multipliers, and the method is the same for most setups.

STEP 1: You should write out what each component of $\nabla f$ and $\nabla g$ are; i.e. compute $\frac{\partial f}{\partial x_i}$ and $\frac{\partial g}{\partial x_i}$ for each $1 \leq i \leq n$. This should just be calculus computations.

STEP 2: By equating $\nabla f = \lambda \nabla g$ you are essentially writing down a system of $n$ equations; one equation for each component of the vectors, i.e. $$\frac{\partial f}{\partial x_i} = \lambda \frac{\partial g}{\partial x_i}$$ You should get equations that look like $$\frac{\sqrt[n]{x_1 x_2 \cdots x_n}}{n x_i} = \lambda \tag{1}$$

STEP 3: Formally solving those equations for each $x_i$ you can now plug in those expressions into $x_1 +x_2 + \ldots + x_n$ to get $$c = \sum_{i=1}^n \frac{\sqrt[n]{x_1 x_2 \cdots x_n}}{n \lambda} = \frac{\sqrt[n]{x_1 x_2 \cdots x_n}}{\lambda}$$ and then solve for $\lambda$ in terms of $c$.

STEP 4: Once you've done that you go back to $(1)$ and find that you can write $x_i = \frac{c}{n}$ for all $i$ and plug the vector $(c/n,c/n,\ldots,c/n)$ into $f$ to get $$f(c/n,c/n,\ldots,c/n) = \sqrt[n]{\frac{c^n}{n^n}} = \frac{c}{n}$$

STEP 5: Lagrange multipliers only gives you potential points for extrema, so you must prove that $c/n$ is indeed a maximum and not some other type of critical point. I will leave this step to you.

• I see; you are considering the $x_1...x_n$ to be variables and not numbers (which makes senses since they are unknown) – Grizzly0111 Oct 17 '15 at 4:09
• This is a nice outline of the steps to solving a constrained extremum problem using Lagrange multipliers. For this particular problem, though, you can take a short cut once you have equation $(1)$: it tells you that $\lambda nx_i=\lambda nx_j$ for all $i,j$. (Note that the numerator is $f$.) That should immediately tell you something about the values of the $x_i$ at the critical point. – amd Oct 17 '15 at 5:02
• @amd Right. That's how I would do it, but I wanted to do the systematic approach. :) – Yeldarbskich Oct 17 '15 at 7:53