The weekly revenue for a product is given by ​R(x)equals138.6 x minus 0.015 x squared​, and the weekly cost is ​C(x)equals12 comma 000 plus 69.3 x minus 0.03 x squared plus 0.00001 x cubed​, where x is the number of units produced and sold.
​(a) How many units will give the maximum​ profit?
​(b) What is the maximum possible​ profit?
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Part 1
​(a) The number of units that will give the maximum profit is

enter your response here.
​(Round to the nearest whole number as​ needed.)

### Step 1: Define the profit function.
The profit function $ P(x) $ is the difference between revenue $ R(x) $ and cost $ C(x) $:

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

Given:
- $ R(x) = 138.6x - 0.015x^2 $
- $ C(x) = 12,000 + 69.3x - 0.03x^2 + 0.00001x^3 $

Substitute the expressions for $ R(x) $ and $ C(x) $ into the profit function:

\[
P(x) = \left( 138.6x - 0.015x^2 \right) - \left( 12,000 + 69.3x - 0.03x^2 + 0.00001x^3 \right)
\]

Simplify the expression:

\[
P(x) = 138.6x - 0.015x^2 - 12,000 - 69.3x + 0.03x^2 - 0.00001x^3
\]

\[
P(x) = (138.6x - 69.3x) + (-0.015x^2 + 0.03x^2) - 12,000 - 0.00001x^3
\]

\[
P(x) = 69.3x + 0.015x^2 - 12,000 - 0.00001x^3
\]

### Step 2: Find the critical points.
To find the number of units that maximize the profit, we take the derivative of $ P(x) $ with respect to $ x $ and set it equal to zero.

\[
P'(x) = \frac{d}{dx} \left( 69.3x + 0.015x^2 - 12,000 - 0.00001x^3 \right)
\]

\[
P'(x) = 69.3 + 2(0.015)x - 3(0.00001)x^2
\]

\[
P'(x) = 69.3 + 0.03x - 0.00003x^2
\]

Set $ P'(x) = 0 $:

\[
69.3 + 0.03x - 0.00003x^2 = 0
\]

### Step 3: Solve the quadratic equation.
Rearrange the equation:

\[
0.00003x^2 - 0.03x - 69.3 = 0
\]

Multiply through by 100,000 to eliminate the decimals:

\[
3x^2 - 3,000x - 6,930,000 = 0
\]

Now solve this quadratic equation using the quadratic formula:

\[
x = \frac{-(-3,000) \pm \sqrt{(-3,000)^2 - 4(3)(-6,930,000)}}{2(3)}
\]

\[
x = \frac{3,000 \pm \sqrt{9,000,000 + 83,160,000}}{6}
\]

\[
x = \frac{3,000 \pm \sqrt{92,160,000}}{6}
\]

\[
x = \frac{3,000 \pm 9,600}{6}
\]

Thus:

\[
x_1 = \frac{3,000 + 9,600}{6} = \frac{12,600}{6} = 2,100
\]

\[
x_2 = \frac{3,000 - 9,600}{6} = \frac{-6,600}{6} = -1,100
\]

Since the number of units produced must be positive, we choose $ x = 2,100 $.

### Step 4: Verify that this is a maximum.
To confirm that $ x = 2,100 $ gives a maximum profit, we take the second derivative of $ P(x) $:

\[
P''(x) = \frac{d}{dx} \left( 69.3 + 0.03x - 0.00003x^2 \right)
\]

\[
P''(x) = 0.03 - 0.00006x
\]

Substitute $ x = 2,100 $ into $ P''(x) $:

\[
P''(2,100) = 0.03 - 0.00006(2,100) = 0.03 - 0.126 = -0.096
\]

Since $ P''(2,100) < 0 $, the profit is maximized at $ x = 2,100 $.

### Step 5: Calculate the maximum profit.
Now substitute $ x = 2,100 $ into the profit function $ P(x) $:

\[
P(2,100) = 69.3(2,100) + 0.015(2,100)^2 - 12,000 - 0.00001(2,100)^3
\]

\[
P(2,100) = 145,530 + 0.015(4,410,000) - 12,000 - 0.00001(9,261,000,000)
\]

\[
P(2,100) = 145,530 + 66,150 - 12,000 - 92,610 = 107,070
\]

Thus, the maximum possible profit is **$107,070**.

### Final Answers:
(a) The number of units that will give the maximum profit is **2,100** units.

(b) The maximum possible profit is **$107,070**.

Would you like more details or clarifications?

---

Here are 5 questions to further expand your understanding:
1. How would changes in the cost function impact the profit-maximizing number of units?
2. What happens to profit if fixed costs increase by 10%?
3. How can you find the break-even point for this product?
4. Can the maximum profit be found if the cubic term is omitted from the cost function?
5. How would taxes or subsidies on production affect the profit function?

**Tip:** Always check the second derivative to ensure a maximum or minimum in optimization problems.

The weekly revenue for a product is given by ​R(x) = 138.6x - 0.015x^2, and the weekly cost is ​C(x) = 12,000 + 69.3x - 0.03x^2 + 0.00001x^3, where x is the number of units produced and sold. ​(a) How many units will give the maximum​ profit? ​(b) What is the maximum possible​ profit?

Learn how to find the number of units that maximizes profit for a product using the given revenue and cost functions. This detailed solution covers calculus concepts like the profit function, first and second derivatives, and optimization techniques. Ideal for college-level students.

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