Advanced Math Test #4

Given that

$\begin{aligned} \frac x{1+x} + \frac {2y}{1+y} & = 1 \\ \frac {x(1+y)+2y(1+x)}{(1+x)(1+y)} & = 1 \\ x + xy + 2y + 2xy & = 1 + x +y+xy \\ y + 2xy & = 1 \\ xy & = \frac 12 (1-y) & \small \blue{\text{Multiply both sides by }y} \\ xy^2 & = \frac 12 \left(y-y^2\right) \\ & = - \frac 12 \left(y-\frac 12\right)^2 + \frac 18 & \small \blue{\text{Since }\left(y-\frac 12\right)^2 \ge 0} \\ \implies A & \le \frac 18 \end{aligned}$

Therefore $a+b = 1+8 = \boxed 9$ .

The constraint can be simplified to read $\begin{aligned} \frac{2y}{1+y} & = \; \frac{1}{1+x} \\ 2(1+x)y & = \; 1+y \\ y & = \; \frac{1}{1+2x} \end{aligned}$ and hence we want to maximize $A \; = \; xy^2 \; = \; \frac{x}{(1 +2x)^2}$ which is maximized when $x=\tfrac12$ , with a maximum value of $\tfrac18$ , making the answer $\boxed{9}$ .

$A_{max}=\dfrac{1}{8}$