Island Water Sports provides rental equipment and instruction for a variety of water sports in a resort town. On one morning, a decision must be made on how many Wildlife Raft Trips and how many Group Sailing Lessons should be scheduled. Each Wildlife Raft Trip requires one captain and one crew person and can accommodate six passengers. The revenue per raft trip is $120. Ten rafts are available, and at least 30 people are on the list for reservations this morning. Each Group Sailing Lesson requires one captain and two crew people for instruction. Two boats are needed for each group. Four students form each group. There are 12 sailboats available, and at least 20 people are on the list for sailing instruction this morning. The revenue per group sailing lesson is $160. The company has 12 captains and 18 crew available this morning. The company would like to maximize the number of customers served while generating at least $1800 in revenue and honoring all reservations. Define the decision variables as follows and formulate the problem as a linear program. Let x be the number of Wildlife raft Trips to be scheduled. Let y be the number of Group Sailing Lessons to be scheduled. The following constraints complete the problem:
120x + 160y ≥ 1800 (1)
x ≤ 10 (2)
2y ≤ 12 (3)
6x ≥ 30 (4)
4y ≥ 20 (5)
x + y ≤ 12 (6)
x + 2y ≤ 18 (7)
1) What does constraint (1) represent?
A) The maximum number of captions who can be deployed.
B) The maximum number of crews who can be deployed.
C) The minimum revenue requirement.
D) The maximum quantity of the rafts available.
E) The maximum quantity of the sailboats available.
F) The demand of raft trips that must be satisfied.
G) The demand of sailing lessons that must be satisfied.
2) What does constraint (4) represent?
A) The maximum number of captions who can be deployed.
B) The maximum number of crews who can be deployed.
C) The minimum revenue requirement.
D) The maximum quantity of the rafts available.
E) The maximum quantity of the sailboats available.
F) The demand of raft trips that must be satisfied.
G) The demand of sailing lessons that must be satisfied.
3) What does constraint (6) represent?
A) The maximum number of captions who can be deployed.
B) The maximum number of crews who can be deployed.
C) The minimum revenue requirement.
D) The maximum quantity of the rafts available.
E) The maximum quantity of the sailboats available.
F) The demand of raft trips that must be satisfied.
G) The demand of sailing lessons that must be satisfied.

1)  C

2) F

3) A

Step-by-step explanation:

Given that :  X = number of Wildlife raft Trips to be scheduled.

while Y = number of Group Sailing Lessons to be scheduled.

1) Constraint (1) : 120x + 160y ≥ 1800

Represents the Minimum revenue requirement

2) Constraint (4) : 6x ≥ 30

Represents The demand of raft trips that must be satisfied.

3) Constraint (6) : x + y ≤ 12

Represents The maximum number of captions who can be deployed.

answer: i believe its b. if you sound out the sentences that sounds wrong.

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).