The monthly profit in dollars from the production and sale of 𝒙 units of a product is
given by the equation 𝑷(𝒙) = −𝟒𝒙𝟐 + 𝟒𝟎𝟎𝒙 − 𝟑𝟔𝟎𝟎
a) (2 points) How many units must be produced and sold to maximize the profit?
b) (2 points) What is the maximum profit?
c) (2 points) What levels of production and sales will give a monthly profit of $6000?
d) (2 points) What levels of production and sales will result in breakeven (zero profit)?

Given the profit function $ P(x) = -4x^2 + 400x - 3600 $, where $ P(x) $ represents the monthly profit in dollars from the production and sale of $ x $ units, let's tackle each part of the question:

### a) How many units must be produced and sold to maximize the profit?

The profit function $ P(x) = -4x^2 + 400x - 3600 $ is a quadratic equation. Since the coefficient of $ x^2 $ is negative ($ -4 $), the parabola opens downward, meaning it has a maximum point.

To find the number of units that maximizes the profit, we need to find the vertex of the parabola. For a quadratic equation in the form $ ax^2 + bx + c $, the x-coordinate of the vertex is given by:

\[
x = -\frac{b}{2a}
\]

In this case, $ a = -4 $ and $ b = 400 $, so:

\[
x = -\frac{400}{2(-4)} = \frac{400}{8} = 50
\]

Thus, the profit is maximized when 50 units are produced and sold.

### b) What is the maximum profit?

Now that we know the profit is maximized when 50 units are produced and sold, we can find the maximum profit by plugging $ x = 50 $ into the profit equation:

\[
P(50) = -4(50)^2 + 400(50) - 3600
\]
\[
P(50) = -4(2500) + 20000 - 3600
\]
\[
P(50) = -10000 + 20000 - 3600 = 6400
\]

Therefore, the maximum profit is **$6400**.

### c) What levels of production and sales will give a monthly profit of $6000?

To find the levels of production that yield a profit of $6000, we set $ P(x) = 6000 $ and solve for $ x $:

\[
6000 = -4x^2 + 400x - 3600
\]

Rearranging the equation:

\[
0 = -4x^2 + 400x - 9600
\]

Now, divide the entire equation by -4 to simplify:

\[
0 = x^2 - 100x + 2400
\]

Solve this quadratic equation using the quadratic formula:

\[
x = \frac{-(-100) \pm \sqrt{(-100)^2 - 4(1)(2400)}}{2(1)}
\]
\[
x = \frac{100 \pm \sqrt{10000 - 9600}}{2}
\]
\[
x = \frac{100 \pm \sqrt{400}}{2}
\]
\[
x = \frac{100 \pm 20}{2}
\]

This gives two possible values for $ x $:

\[
x = \frac{100 + 20}{2} = 60 \quad \text{or} \quad x = \frac{100 - 20}{2} = 40
\]

Thus, a monthly profit of $6000 occurs when **40 or 60 units** are produced and sold.

### d) What levels of production and sales will result in breakeven (zero profit)?

To find the breakeven points, we set $ P(x) = 0 $ and solve for $ x $:

\[
0 = -4x^2 + 400x - 3600
\]

Divide the equation by -4 to simplify:

\[
0 = x^2 - 100x + 900
\]

Solve this quadratic equation using the quadratic formula:

\[
x = \frac{-(-100) \pm \sqrt{(-100)^2 - 4(1)(900)}}{2(1)}
\]
\[
x = \frac{100 \pm \sqrt{10000 - 3600}}{2}
\]
\[
x = \frac{100 \pm \sqrt{6400}}{2}
\]
\[
x = \frac{100 \pm 80}{2}
\]

This gives two possible values for $ x $:

\[
x = \frac{100 + 80}{2} = 90 \quad \text{or} \quad x = \frac{100 - 80}{2} = 10
\]

Thus, the breakeven points occur when **10 or 90 units** are produced and sold.

---

### Summary:
- **a)** Maximum profit is achieved when 50 units are produced and sold.
- **b)** The maximum profit is $6400.
- **c)** A monthly profit of $6000 is achieved when 40 or 60 units are produced and sold.
- **d)** Breakeven occurs when 10 or 90 units are produced and sold.

---

Would you like more details on any step?

### 5 Related Questions:
1. How do you graph the profit function $ P(x) $ and interpret its features?
2. What is the significance of the discriminant in quadratic equations for finding solutions?
3. How can you find the vertex of a quadratic function using completing the square?
4. What are some other methods to solve quadratic equations besides the quadratic formula?
5. How does the second derivative help in confirming the nature of the turning point?

### Tip:
When solving quadratic equations, always check if the equation can be factored before applying the quadratic formula. This can often save time.

The monthly profit in dollars from the production and sale of 𝒙 units of a product is given by the equation 𝑷(𝒙) = −𝟒𝒙𝟐 + 𝟒𝟎𝟎𝒙 − 𝟑𝟔𝟎𝟎. a) How many units must be produced and sold to maximize the profit? b) What is the maximum profit? c) What levels of production and sales will give a monthly profit of $6000? d) What levels of production and sales will result in breakeven (zero profit)?

This page explains how to maximize profit using the quadratic function P(x) = -4x^2 + 400x - 3600. It includes step-by-step solutions for finding the maximum profit, determining break-even points, and solving for specific profit levels. Suitable for Grades 9-11, covering key algebraic concepts and quadratic equations.

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