The three verticies show where a max can occur.
Checking near each one( they are not all integers)
15 glass trays 0 decanters
12 glass trays 3 decanters
1 glass tray 6 decanters
plugging each of these into the profit function gives
So the most profit occus with 6 decanters and 1 glass tray.
for b and the constraint that
find your new vertices and repeat the same process.
I hope this helps.