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Find the quantity of the raw material such that the company has the maximum profit.

Find the quantity of the raw material (x > 1) such that the company has no profit.

### Problem 1: **Finding the quantity of raw material for maximum profit**

We are given the profit function:

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

This is a quadratic function of the form \( P(x) = ax^2 + bx + c \), where \( a = -2 \), \( b = 8 \), and \( c = -6 \).

To find the quantity \( x \) that maximizes the profit, we use the fact that for a quadratic function, the maximum (or minimum) occurs at the vertex. The \( x \)-coordinate of the vertex is given by:

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

Substituting \( a = -2 \) and \( b = 8 \) into the formula:

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

Thus, the quantity of raw material that maximizes profit is **2 units**.

---

### Problem 2: **Finding the quantity of raw material for no profit (P(x) = 0 and x > 1)**

To find when the company has no profit, we set \( P(x) = 0 \) and solve for \( x \):

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

We can solve this quadratic equation using the quadratic formula:

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

For the equation \( -2x^2 + 8x - 6 = 0 \), the values of \( a \), \( b \), and \( c \) are:

- \( a = -2 \)
- \( b = 8 \)
- \( c = -6 \)

Substituting into the quadratic formula:

\[
x = \frac{-8 \pm \sqrt{8^2 - 4(-2)(-6)}}{2(-2)}
\]
\[
x = \frac{-8 \pm \sqrt{64 - 48}}{-4}
\]
\[
x = \frac{-8 \pm \sqrt{16}}{-4}
\]
\[
x = \frac{-8 \pm 4}{-4}
\]

Now, solving for the two possible values of \( x \):

1. \( x_1 = \frac{-8 + 4}{-4} = \frac{-4}{-4} = 1 \)
2. \( x_2 = \frac{-8 - 4}{-4} = \frac{-12}{-4} = 3 \)

Thus, the company has no profit when \( x = 1 \) or \( x = 3 \). Since we are asked for \( x > 1 \), the quantity of raw material is **3 units**.

---

Would you like more details or have any questions? Here are 5 related questions to extend this problem:

1. What is the maximum profit value when the quantity of raw material is 2 units?
2. How would the profit function change if the cost of raw materials increases?
3. Can you derive the break-even point where profit transitions from negative to positive?
4. How would the maximum profit change if the quadratic term's coefficient were less negative?
5. How does the company’s profit function change if a constant subsidy is added?

**Tip**: To quickly find the turning point of a quadratic function, use the vertex formula \( x = -\frac{b}{2a} \).

Find the quantity of the raw material such that the company has the maximum profit. Find the quantity of the raw material (x > 1) such that the company has no profit.

Learn how to find the quantity of raw material that maximizes profit and determine when the profit is zero using a quadratic profit function. This involves using the vertex formula and solving the quadratic equation. Suitable for students in Grades 10-12.

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