Calculating the optimal gym membership price to maximize revenue using a quadratic equation.

Problem Statement:

A gym charges its members \$300 per year to join. With this price, they get 1000 members. For each \$2 increase in the price, they expect to lose 5 members. The task is to determine how much the gym should charge to maximize its revenue and calculate the maximum revenue.

---

Solution:

#### Step 1: Define Variables
- Let \( x \) be the number of \$2 increases in the price.
- The new price per member will be \( 300 + 2x \).
- The number of members will be \( 1000 - 5x \).

#### Step 2: Revenue Formula
Revenue is calculated as:
\[
\text{Revenue} = (\text{Number of Members}) \times (\text{Price per Member})
\]
Substituting the expressions for the number of members and the price per member:
\[
\text{Revenue} = (1000 - 5x)(300 + 2x)
\]

#### Step 3: Expand the Revenue Expression
Expand the product:
\[
\text{Revenue} = (1000 - 5x)(300 + 2x)
\]
\[
= 1000 \cdot 300 + 1000 \cdot 2x - 5x \cdot 300 - 5x \cdot 2x
\]
\[
= 300000 + 2000x - 1500x - 10x^2
\]
\[
= -10x^2 + 500x + 300000
\]

So, the revenue function is:
\[
R(x) = -10x^2 + 500x + 300000
\]

#### Step 4: Maximize the Revenue
The revenue function \( R(x) = -10x^2 + 500x + 300000 \) is a quadratic function in the form \( ax^2 + bx + c \), where \( a = -10 \), \( b = 500 \), and \( c = 300000 \). Since \( a < 0 \), the parabola opens downwards, and the maximum value occurs at the vertex.

The \( x \)-coordinate of the vertex of a parabola given by \( ax^2 + bx + c \) is:
\[
x = -\frac{b}{2a}
\]
Substitute \( a = -10 \) and \( b = 500 \):
\[
x = -\frac{500}{2(-10)} = \frac{500}{20} = 25
\]

#### Step 5: Calculate the Optimal Price
The optimal number of \$2 increases is \( x = 25 \). The new price per member is:
\[
\text{Price} = 300 + 2x = 300 + 2(25) = 300 + 50 = 350
\]

#### Step 6: Calculate the Maximum Revenue
Substitute \( x = 25 \) back into the revenue function:
\[
R(x) = -10x^2 + 500x + 300000
\]
\[
R(25) = -10(25)^2 + 500(25) + 300000
\]
\[
= -10(625) + 12500 + 300000
\]
\[
= -6250 + 12500 + 300000
\]
\[
= 306250
\]

#### Step 7: Verify the Number of Members
When \( x = 25 \), the number of members is:
\[
\text{Members} = 1000 - 5x = 1000 - 5(25) = 1000 - 125 = 875
\]
The revenue is:
\[
\text{Revenue} = 875 \times 350 = 306250
\]
This confirms our calculation.

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Final Answer:

The gym should charge \$350 per year to maximize its revenue. The maximum revenue is \$306,250.

\[
\boxed{350 \text{ and } 306250}
\]