Let C(x) be the cost to produce x batches of widgets, and let R(x) be the revenue in thousands of dollars. R(x)=-x^2+8x, C(x)=x+6. Find the minimum​ break-even quantity.
Using the expressions -x^2+8x and/or x+6, identify an equation to be solved in order to find the minimum​ break-even quantity. Find the maximum revenue.
How can the maximum revenue be​ found? The maximum revenue is... Find the maximum profit.
Let P(x) be the profit in thousands of dollars. Identify an expression in terms of x for P(x). P(x)=

To solve the given problem, let's break it down into steps:

### 1. **Break-even quantity**  
The break-even point occurs when the revenue equals the cost, i.e., $ R(x) = C(x) $. This means we set the two functions equal and solve for $ x $:

\[
R(x) = C(x) \implies -x^2 + 8x = x + 6
\]

This is the equation that must be solved to find the **break-even quantity**.

### 2. **Find the minimum break-even quantity**

To solve $ -x^2 + 8x = x + 6 $:

\[
-x^2 + 8x - x - 6 = 0
\]
\[
-x^2 + 7x - 6 = 0
\]

Multiply through by $ -1 $ to make the coefficient of $ x^2 $ positive:

\[
x^2 - 7x + 6 = 0
\]

Factor the quadratic equation:

\[
(x - 6)(x - 1) = 0
\]

Thus, the solutions are $ x = 6 $ and $ x = 1 $.  
These are the break-even points, so the **minimum break-even quantity** is $ x = 1 $.

### 3. **Find the maximum revenue**

The revenue function is $ R(x) = -x^2 + 8x $. This is a quadratic function, and since the coefficient of $ x^2 $ is negative, the parabola opens downward. The maximum revenue occurs at the vertex of the parabola.

The formula for the $ x $-coordinate of the vertex for a quadratic equation $ ax^2 + bx + c $ is:

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

For $ R(x) = -x^2 + 8x $, we have $ a = -1 $ and $ b = 8 $:

\[
x = \frac{-8}{2(-1)} = 4
\]

Thus, the maximum revenue occurs at $ x = 4 $. To find the value of the maximum revenue, substitute $ x = 4 $ into $ R(x) $:

\[
R(4) = -(4)^2 + 8(4) = -16 + 32 = 16
\]

Thus, the **maximum revenue is 16 thousand dollars** (or $16,000).

### 4. **Find the maximum profit**

Profit $ P(x) $ is defined as revenue minus cost:

\[
P(x) = R(x) - C(x)
\]

Using the given functions $ R(x) = -x^2 + 8x $ and $ C(x) = x + 6 $, we have:

\[
P(x) = (-x^2 + 8x) - (x + 6)
\]
\[
P(x) = -x^2 + 8x - x - 6
\]
\[
P(x) = -x^2 + 7x - 6
\]

This is the expression for the profit function $ P(x) $.

### 5. **Find the maximum profit**

Since $ P(x) = -x^2 + 7x - 6 $ is a quadratic function with a negative $ x^2 $-term, it also has a maximum at its vertex. The $ x $-coordinate of the vertex is:

\[
x = \frac{-7}{2(-1)} = \frac{7}{2} = 3.5
\]

To find the maximum profit, substitute $ x = 3.5 $ into $ P(x) $:

\[
P(3.5) = -(3.5)^2 + 7(3.5) - 6
\]
\[
P(3.5) = -12.25 + 24.5 - 6 = 6.25
\]

Thus, the **maximum profit is 6.25 thousand dollars** (or $6,250).

---

### Summary:

- **Minimum break-even quantity**: $ x = 1 $
- **Maximum revenue**: $16,000
- **Profit function**: $ P(x) = -x^2 + 7x - 6 $
- **Maximum profit**: $6,250

---

Would you like more details on any of the steps?

Here are 5 related questions:
1. What does the second break-even quantity represent in this context?
2. How does the vertex formula help in finding maximum or minimum points in quadratic functions?
3. Can the profit function have more than one maximum? Why or why not?
4. What is the economic significance of the break-even point in business?
5. How would increasing the cost function $ C(x) $ affect the break-even points?

**Tip**: Always check whether the quadratic opens upwards or downwards before determining if it has a maximum or minimum value.

Let C(x) be the cost to produce x batches of widgets, and let R(x) be the revenue in thousands of dollars. R(x) = -x^2 + 8x, C(x) = x + 6. Find the minimum break-even quantity. Using the expressions -x^2 + 8x and/or x + 6, identify an equation to be solved in order to find the minimum break-even quantity. Find the maximum revenue. How can the maximum revenue be found? The maximum revenue is... Find the maximum profit. Let P(x) be the profit in thousands of dollars. Identify an expression in terms of x for P(x).

Learn how to find the break-even quantity, maximum revenue, and maximum profit for a production problem where R(x) = -x^2 + 8x and C(x) = x + 6. This solution covers key quadratic concepts and shows how to use the quadratic formula and vertex method to find break-even points and optimize profit.

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