given the following expression:
$V =\frac{a}{a+x}+\frac{b}{b+y}+\frac{c}{c+z}$
and the following conditions: $\\~\\a+b+c=30, a\neq 0 \\~\\x+y+z=15$
I am supposed to find the maximum possible value of V. (All variables are non-negative integers)
I am not sure how to go about this, the only approach that comes to mind is a brute force solution which I think is probably avoidable.
I also tried to substitute $c$ = $30-(a+b)$ and $z$ = $15-(x+y)$ but didn't really get anywhere. Is there a better way to solve this? Am I missing something obvious here?
Thanks so much in advance
For a more-or-less "brute force" way, use the method of Lagrange multipliers. Maximize the quantity \begin{align} \mathcal{L} &= V \;+\; \lambda_1\left(a+b+c-30\right) \;+\; \lambda_2\left(x+y+z - 15\right)\\ &= \frac{a}{a+x} + \frac{b}{b+y} + \frac{c}{c+z} \;+\; \lambda_1\left(a+b+c-30\right) \;+\; \lambda_2\left(x+y+z - 15\right)\, . \end{align} Taking derivatives with respect to $a$, $b$, $c$, $x$, $y$, and $z$ and setting them equal to zero, we have: \begin{align} \frac{\partial \mathcal{L}}{\partial a} &= \frac{x}{{(a+x)}^2} + \lambda_1 = 0& (1)\\ \frac{\partial \mathcal{L}}{\partial b} &= \frac{y}{{(b+y)}^2} + \lambda_1 = 0& (2)\\ \frac{\partial \mathcal{L}}{\partial c} &= \frac{z}{{(c+z)}^2} + \lambda_1 = 0& (3)\\ \frac{\partial \mathcal{L}}{\partial x} &= -\frac{a}{{(a+x)}^2} + \lambda_2 = 0& (4)\\ \frac{\partial \mathcal{L}}{\partial y} &= -\frac{b}{{(b+y)}^2} + \lambda_2 = 0& (5)\\ \frac{\partial \mathcal{L}}{\partial z} &= -\frac{c}{{(c+z)}^2} + \lambda_2 = 0& (6) \end{align} Combining Eqs. (1) and (4), we can eliminate the quantity $1/{(a+x)}^2$ to yield: \begin{equation} x = -\frac{\lambda_1}{\lambda_2}a \qquad (7) \end{equation} In an entirely analogous way, we can combine Eqs. (2) and (5) to yield \begin{equation} y = -\frac{\lambda_1}{\lambda_2}b \qquad (8) \end{equation} and Eqs. (3) and (6) to yield \begin{equation} z = -\frac{\lambda_1}{\lambda_2}c \qquad (9) \end{equation} Now add together Eqs. (7), (8), and (9): \begin{equation} x+y+z = -\frac{\lambda_1}{\lambda_2}(a+b+c)\quad \longrightarrow\quad 15 = -\frac{\lambda_1}{\lambda_2}\times 30 \quad \longrightarrow\quad -\frac{\lambda_1}{\lambda_2} = \frac{1}{2} \end{equation} Via Eqs. (7), (8), and (9), this then implies that \begin{equation} \frac{x}{a} = \frac{y}{b} = \frac{z}{c} = \frac{1}{2}\, . \end{equation} Finally, rewrite $V$ as \begin{align} V &= \frac{1}{1 + x/a} + \frac{1}{1 + y/b} + \frac{1}{1 + z/c}\\ &= \frac{1}{1 + 1/2} + \frac{1}{1 + 1/2} + \frac{1}{1 + 1/2}\\ &= 2 \end{align} This maximum value is attained when $a = b = c = 10$ and $x = y = z = 5$.