The correct answer is option B.
Let the basic calendars be = x
Let the deluxe calendars be = y
Total calendars are = 72 , so equation becomes:
.............(1)
Cost of 1 basic calendar = $10
So, cost of all basic calendars will be = 10x
Cost if 1 deluxe calendar = $15
So, cost of all deluxe calendars = 15y
As given, the number of deluxe calendars must be greater than or equal to 3 times the number of basic calendars, the equation becomes
y=3x ...............(2)
Putting the value of y from (2) in (1)
![x+y=72](https://img.qammunity.org/2019/formulas/mathematics/college/ylez1mk1uhnfntfz35ix7asewfhxgnmrwg.png)
![x+3x=72](https://img.qammunity.org/2019/formulas/mathematics/college/byv6bwivig5w8qifp9j8p2s96ffm9tg2vr.png)
![4x=72](https://img.qammunity.org/2019/formulas/mathematics/college/xg48x0obiclb8csmdy2vbklktoewdo1e8b.png)
x=18
As, y=3x
y=
![3*18=54](https://img.qammunity.org/2019/formulas/mathematics/college/xrnso1t2mw041ym68vt14ul5edowbiaijb.png)
So we have 18 basic calendars and 54 deluxe calendars.
Cost of 18 basic calendars will be=
![18*10=180](https://img.qammunity.org/2019/formulas/mathematics/college/ebshl90zh1pf3d1bi5fl0r72ffedoya6eo.png)
Cost of 54 deluxe calendars will be =
![54*15=810](https://img.qammunity.org/2019/formulas/mathematics/college/sxrrt9qq8zecw18euen4srsfs3rp02khxu.png)
So total amount is = $990
Hence, the maximum revenue is $990.