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
![\begin{array}{rrcrl}(1) & x + y & = &1740&\\(2) & 5.00x + 7.50y & = & 11 100\\(3)& 5.00x + 5.00y & = & 8700 & \text{Multiplied (1) by 5}\\&2.50 y & = &2400 &\text{Subtracted (3) from (2)}\\(4)&y& = &\mathbf{960} &\text{Divided each side by 2.50}\\& x +960& = &1740& \text{Substituted (4) into (1)}\\& x& = &\mathbf{780}& \\\end{array}\\\text{There are $\large \boxed{\textbf{780 students and 960 non-students}}$}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brrcrl%7D%281%29%20%26%20x%20%2B%20y%20%26%20%3D%20%261740%26%5C%5C%282%29%20%26%205.00x%20%2B%207.50y%20%26%20%3D%20%26%2011%20100%5C%5C%283%29%26%20%205.00x%20%2B%205.00y%20%26%20%3D%20%26%208700%20%26%20%5Ctext%7BMultiplied%20%281%29%20by%205%7D%5C%5C%262.50%20y%20%26%20%3D%20%262400%20%26%5Ctext%7BSubtracted%20%283%29%20from%20%282%29%7D%5C%5C%284%29%26y%26%20%3D%20%26%5Cmathbf%7B960%7D%20%26%5Ctext%7BDivided%20each%20side%20by%202.50%7D%5C%5C%26%20x%20%2B960%26%20%3D%20%261740%26%20%5Ctext%7BSubstituted%20%284%29%20into%20%281%29%7D%5C%5C%26%20x%26%20%3D%20%26%5Cmathbf%7B780%7D%26%20%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BThere%20are%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B780%20students%20and%20960%20non-students%7D%7D%24%7D)
(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.