The cost function for a cartain company is \( c=60 x+300 \) and the revenue is given by \( R=100 x-0,5 x^{2} \). Recall that profit is revenue minus cost, Set up a quadratic equation ond find two values of \( x \) (production level) that will create a profit of \( \leq 300 \). \( x= \)

To find the production levels \( x \) that will create a profit of \( \leq 300 \), we first need to express the profit function. The profit \( P \) is given by the difference between revenue \( R \) and cost \( c \): \[ P = R - c \] Substituting the given functions for revenue and cost: \[ P = (100x - 0.5x^2) - (60x + 300) \] Now, simplify the profit function: \[ P = 100x - 0.5x^2 - 60x - 300 \] \[ P = (100x - 60x) - 0.5x^2 - 300 \] \[ P = 40x - 0.5x^2 - 300 \] Next, we want to find the values of \( x \) such that the profit is less than or equal to 300: \[ 40x - 0.5x^2 - 300 \leq 300 \] Rearranging this inequality gives: \[ 40x - 0.5x^2 - 300 - 300 \leq 0 \] \[ -0.5x^2 + 40x - 600 \leq 0 \] To make it easier to work with, we can multiply the entire inequality by -1 (remembering to flip the inequality sign): \[ 0.5x^2 - 40x + 600 \geq 0 \] Now, we can multiply through by 2 to eliminate the fraction: \[ x^2 - 80x + 1200 \geq 0 \] Next, we will find the roots of the quadratic equation \( x^2 - 80x + 1200 = 0 \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = -80 \), and \( c = 1200 \): \[ x = \frac{-(-80) \pm \sqrt{(-80)^2 - 4 \cdot 1 \cdot 1200}}{2 \cdot 1} \] \[ x = \frac{80 \pm \sqrt{6400 - 4800}}{2} \] \[ x = \frac{80 \pm \sqrt{1600}}{2} \] \[ x = \frac{80 \pm 40}{2} \] Calculating the two possible values for \( x \): 1. \( x = \frac{80 + 40}{2} = \frac{120}{2} = 60 \) 2. \( x = \frac{80 - 40}{2} = \frac{40}{2} = 20 \) Thus, the two values of \( x \) that will create a profit of \( \leq 300 \) are: \[ x = 20 \quad \text{and} \quad x = 60 \] In conclusion, the production levels \( x \) that will create a profit of \( \leq 300 \) are: \[ x = 20 \quad \text{and} \quad x = 60 \]