Can you explain to me (detailed) on how to do this problem? I will give you a brainly if you get this right.
1 answer:
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Answer:
Revenue , Cost and Profit Function
Step-by-step explanation:
Here we are given the Price/Demand Function as
P(x) = 253-2x
which means when the demand of Cat food is x units , the price will be fixed as 253-2x per unit.
Now let us revenue generated from this demand i.e. x units
Revenue = Demand * Price per unit
R(x) = x * (253-2x)
=
Now let us Evaluate the Cost Function
Cost = Variable cost + Fixed Cost
Variable cost = cost per unit * number of units
= 3*x
= 3x
Fixed Cost = 6972 as given in the problem.
Hence
Cost Function C(x) = 3x+6972
Let us now find the Profit Function
Profit = Revenue - Cost
P(x) = R(x) - C(x)
=
=
Now we have to find the quantity at which we attain break even point.
We know that at break even point
Profit = 0
Hence P(x) = 0
now we have to solve the above equation for x
Dividing both sides by -2 we get
Now we have to find the factors of 3486 whose sum is 125. Which comes out to be 42 and 83
Hence we now solve the above quadratic equation using splitting the middle term method .
Hence
Either (x-42) = 0 or (x-83) = 0 therefore
if x-42= 0 ; x=42
if x-83=0 ; x=83
Smallest of which is 42. Hence the number of units at which it attains the break even point is 42.
Answer:
The max. revenue is $57,121.
This happens with 239 passengers.
The price of the ticket is $239 per person.
Step-by-step explanation:
Let the variable "P" denote the number of passengers.
If P=180, then the ticket cost is $298/person. Then the price is reduced by $1 for each additional person.
In general, if P=(180+x), then the ticket cost becomes (298-x) per person.
The revenue can be defined as the ticket cost multplied by the number of passengers:
We can derive R and equal to zero to maximize the function.
The amount of passengers is:
The price of the tickets is
The revenue is: