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To determine the price per football needed to achieve a daily profit of \[tex]$400, we'll follow these steps:
### Step-by-Step Solution
1. Understand the given equation and context:
- The profit equation is \( y = -4x^2 + 80x - 150 \), where \( y \) represents the profit and \( x \) is the price per football in dollars.
- The manager wants to earn a daily profit of \$[/tex]400.
2. Set up the equation with the desired profit:
- Substitute [tex]\( y \)[/tex] with the desired profit of \[tex]$400 in the equation: \[ 400 = -4x^2 + 80x - 150 \] 3. Form a standard quadratic equation: - Rearrange the equation to the standard quadratic form: \[ -4x^2 + 80x - 150 - 400 = 0 \] \[ -4x^2 + 80x - 550 = 0 \] 4. Simplify the quadratic equation: - To make calculations easier, divide the entire equation by -2 (or any common factor, if desired): \[ 2x^2 - 40x + 275 = 0 \] 5. Solve the quadratic equation: - For a quadratic equation of the form \( ax^2 + bx + c = 0 \), use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). - Here, \( a = 2 \), \( b = -40 \), and \( c = 275 \). Plug these values into the quadratic formula: \[ x = \frac{-(-40) \pm \sqrt{(-40)^2 - 4 \cdot 2 \cdot 275}}{2 \cdot 2} \] \[ x = \frac{40 \pm \sqrt{1600 - 2200}}{4} \] \[ x = \frac{40 \pm \sqrt{-600}}{4} \] 6. Solve the discriminant (b² - 4ac): - The discriminant is \( \sqrt{-600} \), which can be expressed as \( \sqrt{600}i \) where \( i \) is the imaginary unit ( \( i = \sqrt{-1} \)). \[ x = \frac{40 \pm 10\sqrt{6}i}{4} \] \[ x = 10 \pm \frac{5\sqrt{6}i}{2} \] ### Conclusion To meet the daily profit goal of \$[/tex]400, the price per football must be:
- [tex]\( x = 10 - \frac{5\sqrt{6}i}{2} \)[/tex]
- [tex]\( x = 10 + \frac{5\sqrt{6}i}{2} \)[/tex]
These solutions indicate complex numbers, suggesting that within the real-number system, there are no real prices [tex]\( x \)[/tex] that will exactly yield a profit of \$400 given by the profit model. However, these complex numbers show the mathematical consistency with the quadratic equation provided.
2. Set up the equation with the desired profit:
- Substitute [tex]\( y \)[/tex] with the desired profit of \[tex]$400 in the equation: \[ 400 = -4x^2 + 80x - 150 \] 3. Form a standard quadratic equation: - Rearrange the equation to the standard quadratic form: \[ -4x^2 + 80x - 150 - 400 = 0 \] \[ -4x^2 + 80x - 550 = 0 \] 4. Simplify the quadratic equation: - To make calculations easier, divide the entire equation by -2 (or any common factor, if desired): \[ 2x^2 - 40x + 275 = 0 \] 5. Solve the quadratic equation: - For a quadratic equation of the form \( ax^2 + bx + c = 0 \), use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). - Here, \( a = 2 \), \( b = -40 \), and \( c = 275 \). Plug these values into the quadratic formula: \[ x = \frac{-(-40) \pm \sqrt{(-40)^2 - 4 \cdot 2 \cdot 275}}{2 \cdot 2} \] \[ x = \frac{40 \pm \sqrt{1600 - 2200}}{4} \] \[ x = \frac{40 \pm \sqrt{-600}}{4} \] 6. Solve the discriminant (b² - 4ac): - The discriminant is \( \sqrt{-600} \), which can be expressed as \( \sqrt{600}i \) where \( i \) is the imaginary unit ( \( i = \sqrt{-1} \)). \[ x = \frac{40 \pm 10\sqrt{6}i}{4} \] \[ x = 10 \pm \frac{5\sqrt{6}i}{2} \] ### Conclusion To meet the daily profit goal of \$[/tex]400, the price per football must be:
- [tex]\( x = 10 - \frac{5\sqrt{6}i}{2} \)[/tex]
- [tex]\( x = 10 + \frac{5\sqrt{6}i}{2} \)[/tex]
These solutions indicate complex numbers, suggesting that within the real-number system, there are no real prices [tex]\( x \)[/tex] that will exactly yield a profit of \$400 given by the profit model. However, these complex numbers show the mathematical consistency with the quadratic equation provided.
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