4. Parking fees at IIUM are RM 5.00 for IIUM students and RM 7.50 for non-IIUM students. At the end of each day, the total number of vehicles parked that day and the gross receipts for the day are
recorded, but the number of vehicles in each category is not. The following table contains the relevant
information for a recent 4-day period:
Day
Vehicles Parked
Gross Receipts
Monday
1,200
RM 7,125
Tuesday
1.550
RM 9,825
Wednesday Thursday
1.740
1,400
RM 11,100 RM 8,650
(a) How many vehicles in each category used the IIUM parking facilities on Wednesday? (1 point]
(b) If 1,200 vehicles are parked in one day, is it possible to take in gross receipts of RM 10,000?
Explain. [1 point]
(c) Describe all possible gross receipts on a day when 1,200 vehicles are parked. [1 point]
(3 points)

(a) 780 students and 960 non-students  

(b) No. The maximum revenue is RM9000 from 1200 non-students.

(c). Revenue is maximum of RM9000 at 1200 non-students, decreasing by RM2.50 per student to a minimum of RM6000 at 1200 students

Step-by-step explanation:

 Let x = IIUM students and

and  y = non-IIUM students

You have two conditions

(a)          x +         y = total vehicles parked

(b) 5.00x + 7.50y = total gross receipts

(a) Wednesday

From your table,  

       x +          y = 1740

5.00x + 7.70y = RM11 100  

Solve the simultaneous equations

(b) Can 1200 vehicles bring in RM10000?

No. Even if all the cars were from non-students, the most you could get is  

1200 × 7.50 = RM9000

(c) Possible combinations for 1200 vehicles  

Revenue = 5.00x + 7.50y = 5.00x + 7.50(1200 -x) = 5.00x + 9000 - 7.50x =  

Revenue = 9000 - 2.50x

The maximum revenue of RM9000 occurs when there are no student cars and 1200 non-student cars.

For each student car that enters and displaces a non-student, the revenue drops by RM2.50.

Finally. when there are 1200 student cars and no non-students, the revenue has dropped to a minimum of RM6000.  

45

step-by-step explanation:

the revenue of the fitness center, y, is the product of two factors: the number of members and the fee paid by each member. there are 320 members now, but number of members goes down by 10 for each $5 increase in fees, so if x is the number of $5 price increases then members is equal to (320 - 10x). since the fee is currently $45, and the fee increases by $5 for every price bump x, the fee is given by (45 + 5x). so, revenue y is given by y = (320 - 10x)(45 + 5x).