The cost function for a certain 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 and find two values of $x$ (production level) that will create a profit of $\$ 300$.
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Answer
Final Answer: The two values of $x$ (production level) that will create a profit of $300 are $x = 60.0$ and $x = 20.0$. So, the final answer is \(x = \boxed{60.0, 20.0}\)
Step 1 ：The profit is given by the revenue minus the cost. So, we have: Profit = Revenue - Cost
Step 2 ：We are given that the profit is $300. So, we can set up the equation as follows: \(300 = (100x - 0.5x^2) - (60x + 300)\)
Step 3 ：This simplifies to: \(300 = 100x - 0.5x^2 - 60x - 300\)
Step 4 ：We can further simplify this to: \(0.5x^2 - 40x + 600 = 0\)
Step 5 ：This is a quadratic equation in the form ax^2 + bx + c = 0. We can solve this equation for x using the quadratic formula: \(x = [-b ± sqrt(b^2 - 4ac)] / (2a)\)
Step 6 ：Let's plug in the values a = 0.5, b = -40, and c = 600 into the quadratic formula to find the values of x.
Step 7 ：Final Answer: The two values of $x$ (production level) that will create a profit of $300 are $x = 60.0$ and $x = 20.0$. So, the final answer is \(x = \boxed{60.0, 20.0}\)
