A fuel station sell, aproximately, 2140 liters of gasoline, per day, in which each liters costs 1,75 dollars. The owner observed that, reducing the price of the liter, occurs an increase in the volume of fuel sold, in the proportion of 20 liters sold per day for each cent reduced in the price of the liter.
A) Obtain an expression that describes the number of liters sold depending on the price of the liter.
B) Calculate the price so that the revenue is maximum.
C) How many liters should be sold, per day, só that the revenue is maximum?
Someone help me please!!
A) To obtain an expression that describes the number of liters sold depending on the price of the liter, we can use the information given in the question.
Let's denote the price of one liter of gasoline as "p" (in dollars) and the number of liters sold per day as "l".
According to the information given, when the price of the liter is reduced by one cent, the volume of fuel sold increases by 20 liters per day. So, we can express this relationship as:
l = 2140 + (20 * (1.75 - p))
The expression above represents the total number of liters sold per day, depending on the price per liter.
B) To calculate the price at which the revenue is maximum, we need to consider both the volume of fuel sold and the price per liter.
The revenue is calculated by multiplying the number of liters sold (l) by the price per liter (p). So, the revenue (R) can be expressed as:
R = l * p
Substituting the expression for l from part A into the revenue equation, we get:
R = (2140 + (20 * (1.75 - p))) * p
To find the price that maximizes the revenue, we can take the derivative of the revenue function with respect to the price (p) and set it equal to zero. Then we can solve the resulting equation for p. However, the previous expression for the revenue function is not in a suitable form for differentiation.
To simplify the expression, we can expand and rearrange the terms:
R = (2140 * p + 20 * p * (1.75 - p))
R = 2140p + 35p - 20p^2
R = -20p^2 + 2175p
Now we can differentiate the revenue function with respect to p:
dR/dp = -40p + 2175
Setting this equal to zero:
-40p + 2175 = 0
Solving for p:
40p = 2175
p = 2175/40
p ≈ 54.375
So, the price that maximizes the revenue is approximately $54.375.
C) To find out how many liters should be sold per day so that the revenue is maximum, we can substitute the value of p calculated in part B back into the expression for l from part A:
l = 2140 + (20 * (1.75 - 54.375))
l = 2140 + (20 * (-52.625))
l = 2140 - 1052.5
l ≈ 1087.5
So, approximately 1087.5 liters should be sold per day to maximize the revenue.