A certain electronics manufacturer found that the marginal cost C to produce x DVD/Blu-ray players can be found using the equation C = 0.03x2 - 7x+600. If the marginal cost were $531, how many DVD/Blu- ray players were produced?



To find how many DVD/Blu-ray players were produced when the marginal cost $ C = 531 $, we can solve the given equation for $ x $:

\[
C = 0.03x^2 - 7x + 600
\]

Given that $ C = 531 $, substitute this value into the equation:

\[
531 = 0.03x^2 - 7x + 600
\]

Now, solve for $ x $.

### Step 1: Subtract 531 from both sides to set the equation to zero:

\[
0.03x^2 - 7x + 600 - 531 = 0
\]

\[
0.03x^2 - 7x + 69 = 0
\]

### Step 2: Multiply through by 100 to eliminate the decimal:

\[
3x^2 - 700x + 6900 = 0
\]

### Step 3: Use the quadratic formula to solve for $ x $. The quadratic formula is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation $ 3x^2 - 700x + 6900 = 0 $, the coefficients are:
- $ a = 3 $
- $ b = -700 $
- $ c = 6900 $

### Step 4: Plug the values into the quadratic formula:

\[
x = \frac{-(-700) \pm \sqrt{(-700)^2 - 4(3)(6900)}}{2(3)}
\]

\[
x = \frac{700 \pm \sqrt{490000 - 82800}}{6}
\]

\[
x = \frac{700 \pm \sqrt{407200}}{6}
\]

\[
x = \frac{700 \pm 638}{6}
\]

### Step 5: Solve for the two possible values of $ x $:

\[
x_1 = \frac{700 + 638}{6} = \frac{1338}{6} = 223
\]

\[
x_2 = \frac{700 - 638}{6} = \frac{62}{6} \approx 10.33
\]

Since $ x $ represents the number of DVD/Blu-ray players produced, we only consider whole numbers. Therefore, the two possible values for $ x $ are:

- $ x = 223 $ players (which is the most likely solution)
- $ x = 10 $ players (if we round 10.33 down)

Thus, the manufacturer produced either 223 or around 10 DVD/Blu-ray players, with 223 being the more realistic solution.

Would you like more details, or do you have any questions?

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Here are 5 related questions for you to consider:
1. What does the term "marginal cost" represent in economic theory?
2. How can the quadratic formula help solve real-world problems?
3. What are some other methods for solving quadratic equations besides the quadratic formula?
4. How would the solution change if the cost function included additional production constraints?
5. How can businesses use marginal cost functions to optimize production?

**Tip**: Always double-check your quadratic solutions by substituting them back into the original equation to verify their correctness!

A certain electronics manufacturer found that the marginal cost C to produce x DVD/Blu-ray players can be found using the equation C = 0.03x^2 - 7x + 600. If the marginal cost were $531, how many DVD/Blu-ray players were produced?

This solution explains how to determine the number of DVD/Blu-ray players produced when the marginal cost is given by the quadratic equation C = 0.03x^2 - 7x + 600, with a marginal cost of $531. It covers algebraic concepts like solving quadratic equations and applying the quadratic formula. Suitable for students in Grades 9-12.

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