jmc

We solve for $y$ in terms of $x$: $$y=\frac{100-5x}{3}.$$ Then we express $xy$ in terms of $x$: $$xy=x\frac{100-5x}{3}=\frac{100x-5x^2}{3}=-\frac{5}{3}{x}^2+\frac{100}{3}x.$$ The graph of this expression is a parabola facing downwards. The greatest possible value of $xy$ occurs at the vertex of this parabola, which occurs when $x=\frac{-100/3}{2\cdot -5/3}=10$. Then, $$xy=10\cdot \frac{50}{3}=\frac{500}{3}.$$ However, this is not an integer. So, we test the two closest integer values of $x$: $x=9$ and $x=11$, to see if either of these yield integer values for $y$.

When $x=9$, $y=\frac{55}{3}$, which is not an integer. When $x=11$, $y=\frac{45}{3}=15$, which is an integer. In this case, $$xy=11\cdot 15=\boxed{165}.$$