DETAILS MY NOTES OSCOLALG1 2.5 .288 . The cost function for a certain company is \( C=40 x+500 \) 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 and find two values of \( x \) (production level) that will create a profit of \( \$ 500 \). \( x=\square \)

To find the production levels that yield a profit of $500, we first set up the profit function \( P(x) = R(x) - C(x) \). From the given equations, this becomes: \[ P(x) = (100x - 0.5x^2) - (40x + 500) = 60x - 0.5x^2 - 500 \] Setting the profit to $500: \[ 60x - 0.5x^2 - 500 = 500 \] This simplifies to: \[ -0.5x^2 + 60x - 1000 = 0 \] Multiplying through by -2 to eliminate the fraction results in: \[ x^2 - 120x + 2000 = 0 \] Now, use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -120, c = 2000 \): Calculating the discriminant: \[ b^2 - 4ac = (-120)^2 - 4 \times 1 \times 2000 = 14400 - 8000 = 6400 \] Thus, \( x = \frac{120 \pm \sqrt{6400}}{2} \): \[ x = \frac{120 \pm 80}{2} \] This gives us two solutions: 1. \( x = \frac{200}{2} = 100 \) 2. \( x = \frac{40}{2} = 20 \) So, the production levels that create a profit of $500 are: \( x = 100 \) and \( x = 20 \).