Explanation:

a+b+c=30 and ab+bc+ca=192
=> b(a+c)+ca=192
b=30-(a+c)
=> [30-(a+c)](a+c)+ac=192
=> 30(a+c)−(a+c)²+ca=192
=> 30a+30c−a²−c²−ca=192
=> c²+(a−30)c+(a²−30a+192)=0
Now, c is real. This means that discriminant of the above equation in c should be ≥0
=> (a−30)²−4(a²−30a+192)≥0
=>a²+900−60a−4a²+120a−4X192≥0
=> 3a²−60a−132≤0
=> a²−20a−44≤0
=> (a−22)(a+2)≤0
a+2≤0          =>a≤−2a+2≤0=a≤−2, which is not possible
Or a−22≤0    =>a≤22   a−22≤0=a≤22 and
Thus, max possible value is 22.