Raymond B. answered 01/25/21
Math, microeconomics or criminal justice
Let L = number of larger 30 foot floats, let S = number of smaller 15 foot floats
the parade length = 30L + 15S = 10(L+S-1) = 30L+15S + 10L+10S -1 = 40L+25S -1
P = 40L+25S -1
150<40L+25S-1<200
151<40L+25S < 201
151/5< 8L + 5S < 201/5 or
30.2 < 8L + 5S < 40.2
L<5, S<8
0<600L + 300S < 2500
0< 6L+3S < 25
L<4, S<8
the cost constraint limited large floats to 4 instead of 5, without the cost constraints.
small floats are no more than 8, regardless of the cost constraints.
8(300)=2400, but a 9th small float would make 2700>budget allowed
4(600) =2400, but a 5th large float would make 3000> budget
6L+3S<25
either L=0, S=8
L=1, S=6
L=2, S= 4
L=3, S=2
or L=4, S=0
5 possibilities that meet the budget restraints and parade length maximums
but 30.2 < 8L:+5S allows for four more possibilities
L=0, S=7
L=1, S=5
L=2, S =3
L=3, S= 2
any lesser combinations would be too short for the parade length minimum