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The problem presented in the image involves profit calculations based on a revenue function $ R(x) $ and cost function $ C(x) $, where:

- $ R(x) = 1.6x^2 $
- $ C(x) = 0.0002x^3 + 8x + 100 $

We are asked to:
- (a) Find the profit function $ P(x) = R(x) - C(x) $.
- (b) Find the profit when 14 hundred cellphones are sold (which means $ x = 14 $).
- (c) Interpret $ P(14) $.

### (a) Find the profit function $ P(x) $:
The profit function $ P(x) $ is defined as the revenue function minus the cost function:

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

Substitute the given expressions for $ R(x) $ and $ C(x) $:

\[
P(x) = 1.6x^2 - \left(0.0002x^3 + 8x + 100\right)
\]

Simplifying:

\[
P(x) = 1.6x^2 - 0.0002x^3 - 8x - 100
\]

### (b) Find the profit when $ x = 14 $:
To find the profit when 14 hundred cellphones are sold, substitute $ x = 14 $ into the profit function $ P(x) $.

\[
P(14) = 1.6(14)^2 - 0.0002(14)^3 - 8(14) - 100
\]

I'll compute this value for you:

\[
P(14) = 1.6(196) - 0.0002(2744) - 112 - 100
\]

Now, calculating the numerical values:

\[
P(14) = 313.6 - 0.5488 - 112 - 100
\]

\[
P(14) \approx 101.0512
\]

### (c) Interpretation of $ P(14) $:
The value $ P(14) \approx 101.05 $ represents the profit (in hundreds of dollars) when 14 hundred cellphones (or 1,400 cellphones) are sold. This means that the profit is approximately \$10,105.

Let me know if you'd like further details or have questions!

Here are 5 questions to expand your knowledge:
1. How does the profit function change if the cost function has a higher degree term?
2. How would the revenue function impact profit if it had a cubic term instead of a quadratic one?
3. What is the significance of the coefficient of the cubic term in the cost function?
4. How do you interpret a negative profit value?
5. Can you calculate the break-even point for this scenario?

**Tip**: When interpreting results from word problems, always remember to include the units and context—like how many items are sold and the total dollar value of profit.

Suppose that the revenue R, in dollars, from selling cell phones, in hundreds, is R(x) = 1.6x^2. The cost C, in dollars, from selling x cell phones, in hundreds, is C(x) = 0.0002x^3 + 8x + 100. (a) Find the profit function P(x). (b) Find the profit if 14 hundred cellphones are sold. (c) Interpret P(14).

Explore how to calculate profit using revenue and cost functions. This problem involves finding the profit function and evaluating it when 1,400 cellphones are sold. The problem is ideal for students in Grades 10-12 and covers algebraic manipulation of polynomials and profit interpretation.

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