The equation that we can create from this situation is:
i = (190 – 5 x) * (29 + x)
where i is the income and x is the increase in daily rate
Expanding the equation:
i = 5510 + 190x – 145x - 5x^2
i = -5x^2 + 45x + 5510
Taking the 1st derivative:
di/dx = -10x +45
Set to zero to get the maxima:
-10x + 45 = 0
x = 4.5
So the cars should be rented at:
29 + x = 33.5 dollars per day
The maximum income is:
i = (190 – 5*4.5) * (33.5)
i = 5,611.25 dollars
Answer:
c im pretty sure
Step-by-step explanation:
Let "a" and "s" represent the costs of advance and same-day tickets, respectively. Your problem statement gives you two relations.
.. a + s = 35 . . . . . the combined cost of one of each is 35
.. 15a +40s = 900 . . total paid for this combination of tickets was 900
There are many ways to solve these equations. You've probably been introduced to "substitution" and "elimination" (or "addition"). Using substitution for "a", we have
.. a = 35 -s
.. 15(35 -s) +40s = 900 . . substitute for "a"
.. 25s +525 = 900 . . . . . . . simplify
.. 25s = 375 . . . . . . . . . . . .subtract 525
.. s = 15 . . . . . . . . . . . . . . .divide by 25
Then
.. a = 35 -15 = 20
The price of an advance ticket was 20.
The price of a same-day ticket was 15.