Maxed

First we will prove next lemma:
If $a,b,c$ are real numbers and $\frac{a}{b}=\frac{b}{c}=\frac{c}{a}$ then $a=b=c$ .
$\frac{a}{b}=\frac{b}{c}$
$\frac{a}{b} \times b^{2}c=\frac{b}{c} \times b^{2}c$
$abc=b^{3}$
On the same way we get $abc=a^{3}$ and $abc=c^{3}$ , so
$a^{3}=b^{3}=c^{3}$
Because of a,b,c are real numbers we have $a=b=c$ and the lemma is proven.
$\frac{x}{3y}=\frac{y}{2x-5y}=\frac{6x-15y}{x}$
$\frac{x}{3y}=\frac{3y}{3(2x-5y}=\frac{6x-15y}{x}$
$\frac{x}{3y}=\frac{3y}{6x-15y}=\frac{6x-15y}{x}$ (1)
From (1) and lemma for $a=x$ , $b=3y$ and $c=6x-15y$ we have $x=3y$ (2)
From (2) we have $-4x^{2}+36y-8=-4(3y)^{2}+36y-8=-36y^{2}+36y-8=-((6y)^{2}-2 \times 6y \times 3 +3^{2})+1=-(6y-3)^{2}+1$
This polynomial will have max value if $y=\frac{1}{2}$ , so $x+y=3y+y=4y=4 \times \frac{1}{2}=\boxed{2}$ ((2) implies x+y=4y)