GASOLINE

A gas station faces continuous price changes for its best-selling fuel type: Euro 95. The

owner of the station has found out that the relation between the number of liters Q sold per

day, and the price p in euros per liter, is given by

Q = 1000p2 − 6000p + 12 750

The relationship between the number of liters \( Q \) sold per day and the price per liter \( p \) for Euro 95 gasoline is given by:

\[
Q = 1000p^2 - 6000p + 12750.
\]

### Analyzing the Equation

This is a quadratic equation in terms of \( p \), which means it represents a parabola. Here's what each term represents:

- \( 1000p^2 \): This term causes the parabola to open upwards (since the coefficient is positive). It suggests that as the price increases significantly, the sales will drop sharply.
- \( -6000p \): This linear term suggests that as the price per liter \( p \) increases moderately, the sales decrease at a somewhat steady rate.
- \( +12750 \): This constant represents the base number of liters sold when the price effects are not considered.

### Step 1: Find the Critical Point (Optimal Price)

To find the price that maximizes or minimizes the number of liters sold, we need to calculate the vertex of this quadratic function. The price \( p \) that corresponds to the vertex can be found using the formula for the vertex of a parabola:

\[
p = -\frac{b}{2a},
\]

where \( a = 1000 \) and \( b = -6000 \) (from the equation \( 1000p^2 - 6000p + 12750 \)).

Let's calculate the optimal price \( p \):

\[
p = -\frac{-6000}{2 \times 1000} = \frac{6000}{2000} = 3.
\]

So, the optimal price is \( p = 3 \) euros per liter.

### Step 2: Determine the Number of Liters Sold at this Price

Now that we know the price that maximizes or minimizes the number of liters sold is \( p = 3 \), let's substitute \( p = 3 \) into the equation for \( Q \):

\[
Q = 1000(3)^2 - 6000(3) + 12750.
\]

Simplifying:

\[
Q = 1000(9) - 6000(3) + 12750,
\]
\[
Q = 9000 - 18000 + 12750,
\]
\[
Q = 9000.
\]

Thus, at the optimal price of \( p = 3 \) euros per liter, the gas station sells 9,000 liters of fuel per day.

### Step 3: Behavior at Higher and Lower Prices

- When the price is lower than 3 euros per liter, the station will sell more than 9,000 liters but eventually, sales will decrease as prices approach zero due to factors like demand saturation.


A gas station faces continuous price changes for its best-selling fuel type: Euro 95. The owner of the station has found out that the relation between the number of liters Q sold per day, and the price p in euros per liter, is given by Q = 1000p^2 − 6000p + 12750.

Discover how to optimize fuel sales at a gas station using a quadratic equation to relate the number of liters sold per day to the price per liter. The problem focuses on algebra, quadratic equations, and optimization. It is suitable for high school students in Grades 9-12.

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