Solve the equation using the multiplication property of equality and the reciprocal of
2 answers:
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Answer:
(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.
- f(x)=ln(x){Attachment 1}
- g(x)=-5ln(x){Attachment 2}
Option C and B are correct
