Let $S_m = (x_1, x_2, \dots , x_m)$ be the $m$-tuple of positive real numbers with $x_1 + x_2 + \cdots + x_m = m$ for which $P_m = x_1 \cdot x_2^2 \cdot \cdots \cdot x_m^m$ is maximised.
For example, it can be verified that $\lfloor P_{10}\rfloor = 4112$ ($\lfloor \, \rfloor$ is the integer part function).
Find $\sum \lfloor P_m \rfloor$ for $2 \le m \le 15$.
This problem requires knowledge of calculus and the method of Lagrange multipliers, which is used to solve optimization problems with constraints, as in this case.
The Lagrangian function of the problem is
L = $x_1\cdot log(x_1) + 2x_2\cdot log(x_2) + \cdots + mx_m\cdot log(x_m) – \lambda (x_1+x_2+\cdots +x_m-m)$
where $\lambda$ is a Lagrange multiplier. The partial derivatives of L with respect to the variables $x_i$ and $\lambda$ are equated to zero:
$\frac{dL}{dx_i}$ = $log(x_i) + (1+ log(x_i)) – \lambda = 0$ (for $i = 1, 2, …, m$) and
$\frac{dL}{d\lambda}$ = $x_1+x_2+\cdots +x_m-m = 0$ .
From the first equation above, you will find that
$x_i$ = $e^{\lambda-1}$ for all $i = 1, 2, …, m$.
Substitute $x_i$ from this equation into the second equation, you have
$m \cdot e^{\lambda-1} = m$, which gives $\lambda = 1$.
So, $x_i=e^{\lambda-1}=1$ for all $i = 1, 2, …, m$, and hence
$P_m = x_1 \cdot x_2^2 \cdot \cdots \cdot x_m^m = 1^1 \cdot 1^2\cdot \cdots \cdot 1^m = 1$.
We can conclude that $P_m = 1$ for all $m = 1, 2, …, 15$, and hence the sum $\sum \lfloor P_m \rfloor$ for $2 \le m \le 15$ equals to 14 (as the floor of 1 is 1 and we sum it from 2 to 15, so 15-2+1 = 14 times).
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