Answer:
$80
Step-by-step explanation:
Let the number of hours required to make a mailbox = x
Let the number of hours required to make a toy = y
Each mailbox requires 1 hour of work from Carlo and 4 hours from Anita.
Each toy requires 1 hour of work from Carlo and 1 hour from Anita.
The table below summarizes the information for ease of understanding.
![\left|\begin{array}{c|c|c|c}&$Mailbox(x)&$Toy(y)&$Maximum Number of Hours\\--&--&--&------------\\$Carlo&1&1&12\\$Anita&4&1&24\end{array}\right|](https://tex.z-dn.net/?f=%5Cleft%7C%5Cbegin%7Barray%7D%7Bc%7Cc%7Cc%7Cc%7D%26%24Mailbox%28x%29%26%24Toy%28y%29%26%24Maximum%20Number%20of%20Hours%5C%5C--%26--%26--%26------------%5C%5C%24Carlo%261%261%2612%5C%5C%24Anita%264%261%2624%5Cend%7Barray%7D%5Cright%7C)
We have the constraints:
![x+y \leq 12\\4x+y \leq 24\\x \geq 0\\y \geq 0](https://tex.z-dn.net/?f=x%2By%20%5Cleq%2012%5C%5C4x%2By%20%5Cleq%2024%5C%5Cx%20%5Cgeq%200%5C%5Cy%20%5Cgeq%200)
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
![x+y \leq 12\\4x+y \leq 24\\x \geq 0\\y \geq 0](https://tex.z-dn.net/?f=x%2By%20%5Cleq%2012%5C%5C4x%2By%20%5Cleq%2024%5C%5Cx%20%5Cgeq%200%5C%5Cy%20%5Cgeq%200)
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.