12. (2 points) The unit price \( p \) is given with the equation \( p(x)=10-x \) where \( x \) is the demand quantity for this product. The fixed production cost is 10 kn and the variable cost is 3 kn per product unit. Determine all the values of \( x \) for which the profit is positive.

To determine the values of \( x \) for which the profit is positive, we start by calculating the revenue, cost, and profit. 1. **Revenue**: The revenue \( R \) is given by the unit price times the quantity demanded: \[ R(x) = p(x) \times x = (10 - x) \times x = 10x - x^2. \] 2. **Cost**: The total cost \( C \) consists of fixed and variable costs: \[ C(x) = \text{Fixed Cost} + \text{Variable Cost} \times x = 10 + 3x. \] 3. **Profit**: The profit \( P \) is given by revenue minus cost: \[ P(x) = R(x) - C(x) = (10x - x^2) - (10 + 3x) = -x^2 + 7x - 10. \] To find when the profit is positive, we need to solve the inequality: \[ -x^2 + 7x - 10 > 0. \] 4. **Factoring**: First, we find the roots of the equation \( -x^2 + 7x - 10 = 0 \): Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = -1, b = 7, c = -10 \). \[ x = \frac{-7 \pm \sqrt{7^2 - 4 \times -1 \times -10}}{2 \times -1} = \frac{-7 \pm \sqrt{49 - 40}}{-2} = \frac{-7 \pm 3}{-2}. \] The roots are \[ x_1 = \frac{-4}{-2} = 2 \quad \text{and} \quad x_2 = \frac{-10}{-2} = 5. \] 5. **Testing intervals**: We now test the intervals \( (-\infty, 2) \), \( (2, 5) \), and \( (5, \infty) \): - For \( x < 2 \) (take \( x = 0 \)): \( P(0) = -10 \) (negative profit). - For \( 2 < x < 5 \) (take \( x = 3 \)): \( P(3) = -3 + 21 - 10 = 8 \) (positive profit). - For \( x > 5 \) (take \( x = 6 \)): \( P(6) = -36 + 42 - 10 = -4 \) (negative profit). Thus, the profit is positive for the interval \( 2 < x < 5 \). The values of \( x \) for which the profit is positive are \( (2, 5) \).