Option A
The restaurant Manager can afford at most 10 employees for the day
<em><u>Solution:</u></em>
Given that restaurant manager can spend at most $600 a day for operating costs and payroll
It costs $100 each day to operate the bank and $50 dollars a day for each employee
The given inequality is:
![50x + 100\leq 600](https://tex.z-dn.net/?f=50x%20%2B%20100%5Cleq%20600)
Where , "x" is the number of employees per day
Let us solve the inequality for "x"
![50x + 100\leq 600](https://tex.z-dn.net/?f=50x%20%2B%20100%5Cleq%20600)
Add -100 on both sides of inequality
![50x + 100 - 100\leq 600 - 100\\\\50x\leq 500](https://tex.z-dn.net/?f=50x%20%2B%20100%20-%20100%5Cleq%20600%20-%20100%5C%5C%5C%5C50x%5Cleq%20500)
Divide by 50 on both sides of inequality
![\frac{50x}{50}\leq \frac{500}{50}\\\\x\leq 10](https://tex.z-dn.net/?f=%5Cfrac%7B50x%7D%7B50%7D%5Cleq%20%5Cfrac%7B500%7D%7B50%7D%5C%5C%5C%5Cx%5Cleq%2010)
Hence the restaurant Manager can afford at most 10 employees for the day
Thus option A is correct