Answer:
$80
Step-by-step explanation:
Let the number of hours required to make a mailbox = x
Let the number of hours required to make a toy = y
Each mailbox requires 1 hour of work from Carlo and 4 hours from Anita.
Each toy requires 1 hour of work from Carlo and 1 hour from Anita.
The table below summarizes the information for ease of understanding.

We have the constraints:

Each mailbox sells for $10 and each toy sells for $5.
Therefore, Revenue, R(x,y)=10x+5y
The given problem is to:
Maximize, R(x,y)=10x+5y
Subject to the constraints

The graph is plotted and attached below.
From the graph, the feasible region are:
(0,0), (6,0), (4,8) and (0,12)
At (6,0), 10x+5y=10(6)+5(0)=60
At (4,8), 10(4)+5(8)=80
At (0,12), 10(0)+5(12)=60
The maximum revenue occurs when they use 4 hours on mailboxes and 8 hours on toys.
The maximum possible revenue is $80.