Find z.

$x+y+z=36$

$x+y=36-z$

Squaring both sides, we get

$x^2+2xy+y^2=36^2-72z+z^2$

$14+2=1296-72z+z^2$

$z^2-72z+1280=0$

Using the quadratic formula to solve for $z$ , we get two values

$z=40$ and $z=32$

Since $x+y>1$ , we must check by substituting.

When $z=32$ :

$x+y+z=36$

$x+y=36-z$

$x+y=36-32=4$

$x+y>4$

When $z+40$

$x+y+z=40$

$x+y=36-z$

$x+y=36-40=-4$

$x+y<4$

So, $\boxed{z=32}$ .