Answer:
- r = 12.5p(32 -p)
- $16 per ticket
- $3200 maximum revenue
Step-by-step explanation:
The number of tickets sold (q) at some price p is apparently ...
q = 150 + 25(20 -p)/2 = 150 +250 -12.5p
q = 12.5(32 -p)
The revenue is the product of the price and the number of tickets sold:
r = pq
r = 12.5p(32 -p) . . . . revenue equation
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The maximum of revenue will be on the line of symmetry of this quadratic function, which is halfway between the zeros at p=0 and p=32. Revenue will be maximized when ...
p = (0 +32)/2 = 16
The theater should charge $16 per ticket.
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Maximum revenue will be found by using the above revenue function with p=16.
r = 12.5(16)(32 -16) = $3200 . . . . maximum revenue
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<em>Additional comment</em>
The number of tickets sold at $16 will be ...
q = 12.5(32 -16) = 200
It might also be noted that if there are variable costs involved, maximum revenue may not correspond to maximum profit.
At the start, the tank contains A(0) = 50 g of salt.
Salt flows in at a rate of
(1 g/L) * (5 L/min) = 5 g/min
and flows out at a rate of
(A(t)/200 g/L) * (5 L/min) = A(t)/40 g/min
so that the amount of salt in the tank at time t changes according to
A'(t) = 5 - A(t)/40
Solve the ODE for A(t):
A'(t) + A(t)/40 = 5
e^(t/40) A'(t) + e^(t/40)/40 A(t) = 5e^(t/40)
(e^(t/40) A(t))' = 5e^(t/40)
e^(t/40) A(t) = 200e^(t/40) + C
A(t) = 200 + Ce^(-t/40)
Given that A(0) = 50, we find
50 = 200 + C ==> C = -150
so that the amount of salt in the tank at time t is
A(t) = 200 - 150 e^(-t/40)
Answer: The store must sell 40 bikes.
Step-by-step explanation:
y=60x+2400
y=120x
120x=60x+2400
-60x on both sides
60x=2400
divide 60 on both sides
2400/60=40
x=40
Answer:
7/30
Step-by-step explanation:
or 7 out of thirty.
plz correct me if im wrong, ill fix it
It would be below the x-axis
