Price of lunch of each, P = $ 18.45/3 = $ 6.15 .
Now, number of people that can eat the barbecue at a budget of $1850.00 is:
[tex]n=\dfrac{1850}{6.15}\\\\n=300.81[/tex]
Therefore, Mr. Gaines afford maximum 300 people to feed on a budget of $1850.00 .
Hence, this is the required solution.