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 values of \( x \) that will create a profit of $500, we need to set up a quadratic equation based on the profit function. The profit function is given by the difference between the revenue and the cost: \[ \text{Profit} = \text{Revenue} - \text{Cost} \] Given: - Cost function: \( C = 40x + 500 \) - Revenue function: \( R = 100x - 0.5x^2 \) The profit function is: \[ \text{Profit} = R - C = (100x - 0.5x^2) - (40x + 500) \] Simplifying the profit function: \[ \text{Profit} = 100x - 0.5x^2 - 40x - 500 \] \[ \text{Profit} = -0.5x^2 + 60x - 500 \] To find the values of \( x \) that will create a profit of $500, we set the profit function equal to $500 and solve for \( x \): \[ -0.5x^2 + 60x - 500 = 500 \] \[ -0.5x^2 + 60x - 1000 = 0 \] Now, we have a quadratic equation in the form \( ax^2 + bx + c = 0 \), where \( a = -0.5 \), \( b = 60 \), and \( c = -1000 \). We can solve this quadratic equation to find the values of \( x \) that will create a profit of $500. Solve the quadratic equation by following steps: - step0: Solve by factoring: \(-0.5x^{2}+60x-1000=0\) - step1: Calculate: \(-\frac{1}{2}x^{2}+60x-1000=0\) - step2: Factor the expression: \(\frac{1}{2}\left(-x+100\right)\left(x-20\right)=0\) - step3: Divide the terms: \(\left(-x+100\right)\left(x-20\right)=0\) - step4: Separate into possible cases: \(\begin{align}&-x+100=0\\&x-20=0\end{align}\) - step5: Solve the equation: \(\begin{align}&x=100\\&x=20\end{align}\) - step6: Rewrite: \(x_{1}=20,x_{2}=100\) The values of \( x \) that will create a profit of $500 are \( x = 20 \) and \( x = 100 \).