What combination of $x$ and $y$ is a feasible solution that minimizes the value of the objective function?

Min $Z=3x+15y$

$(1)2x+4y12$

$(2)5x+2y10$

$x=0$;$y=3$

$x=0$;$y=5$

$x=5$;$y=0$

$x=6$;$y=0$

$x=4$;$y=1$

none of the above