Question 5 (6 points) 3. Let \( P(x)=-0.3 x^{2}+60 x-500 \) be a profit function for a cable television company. It estimates \( x \) as the number of subscribers in thousands and the profit is measured in thousands of dollars. (Use algebra to solve all the problems. To receive credit, all work must be shown) \[ \begin{array}{l}\text { a. A break-even point is where profit is zero. Use algebra to find the break-even points. Give } \\ \text { an interpretation of what this means for the cable company. (6 points) }\end{array} \]
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To find the break-even points, we set the profit function \( P(x) \) equal to zero: \[ -0.3x^2 + 60x - 500 = 0 \] To solve this quadratic equation, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -0.3 \), \( b = 60 \), and \( c = -500 \). First, calculate the discriminant: \[ b^2 - 4ac = 60^2 - 4(-0.3)(-500) = 3600 - 600 = 3000 \] Now, we can apply the quadratic formula: \[ x = \frac{-60 \pm \sqrt{3000}}{2(-0.3)} = \frac{-60 \pm 10\sqrt{30}}{-0.6} = \frac{60 \mp 10\sqrt{30}}{0.6} \] Calculating the two potential break-even points: \[ x = 100 \mp \frac{10\sqrt{30}}{0.6} \approx 100 \mp 8.16 \] This gives us: \[ x \approx 91.84 \quad \text{or} \quad x \approx 108.16 \] Thus, the break-even points are approximately \( x \approx 91.84 \) thousand subscribers and \( x \approx 108.16 \) thousand subscribers. **Interpretation:** The cable company needs between about 91,840 and 108,160 subscribers to cover its costs and not incur a loss. Below this subscriber range, the company will operate at a loss, while above it, it will begin to make a profit.
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