- Equate the revenue function \(R(x) = -\frac{1}{2} x^2 + 20x - 15\) to the expense function \(E(x) = \frac{1}{4} x^2 + 5x + 25\).
- Simplify to the quadratic form \(3x^2 - 60x + 160 = 0\).
- Solve using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
- The break-even quantities are \(\boxed{x = 3.17 \text{ and } x = 16.83}\).
Explanation
- Setting Up the Equation
To find the break-even points for TechGadget, we need to determine the values of \(x\) where the revenue equals the expenses. This occurs when \(R(x) = E(x)\). Given the functions:
\[R(x) = -\frac{1}{2} x^2 + 20x - 15\]
\[E(x) = \frac{1}{4} x^2 + 5x + 25\]
We set these equations equal to each other to solve for \(x\):
\[-\frac{1}{2} x^2 + 20x - 15 = \frac{1}{4} x^2 + 5x + 25\]
- Simplifying the Equation
To solve the equation \(-\frac{1}{2} x^2 + 20x - 15 = \frac{1}{4} x^2 + 5x + 25\), we first collect all terms on one side:
\[0 = \left(\frac{1}{4} + \frac{1}{2}\right) x^2 + (5 - 20) x + (25 + 15)\]
Simplifying the coefficients, we have:
\[0 = \frac{3}{4} x^2 - 15x + 40\]
To clear the fraction, multiply the entire equation by 4:
\[3x^2 - 60x + 160 = 0\]
- Solving the Quadratic Equation
We solve the quadratic equation \(3x^2 - 60x + 160 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -60\), and \(c = 160\):
\[x = \frac{60 \pm \sqrt{(-60)^2 - 4(3)(160)}}{2(3)}\]
\[x = \frac{60 \pm \sqrt{3600 - 1920}}{6}\]
\[x = \frac{60 \pm \sqrt{1680}}{6}\]
Evaluating the square root, \(\sqrt{1680} \approx 40.9878\). Thus:
\[x_1 = \frac{60 - 40.9878}{6} \approx 3.1687\]
\[x_2 = \frac{60 + 40.9878}{6} \approx 16.8313\]
Rounding to two decimal places, we find \(x \approx 3.17\) and \(x \approx 16.83\).
- Conclusion
Comparing our result with the provided options, we see that the break-even quantities are approximately \(x = 3.17\) and \(x = 16.83\). This matches the first option. Therefore, the quantities at which TechGadget breaks even are \(\boxed{x = 3.17 \text{ and } x = 16.83}\).
Examples
In business management, finding the break-even point is crucial for understanding when a product line becomes profitable. If a company knows their revenue function \(R(x)\) and their cost function \(E(x)\), they can use algebra to calculate the minimum and maximum quantity of units they need to sell to avoid a loss. This analysis helps in setting production targets and pricing strategies to ensure the company remains financially healthy. For example, if a company makes custom phone cases, calculating these points tells them exactly how many units must be sold to cover all manufacturing expenses and begin turning a profit.