Subjects algebra

Profit Quadratic 8029Ad

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1. **State the problem:** We are given a quadratic equation for profit $P$ in terms of $b$, the number of skateboards (in thousands): $$P = -2b^2 + 14b - 20$$ We want to analyze this quadratic function. 2. **Formula and rules:** A quadratic function is generally written as: $$P = ax^2 + bx + c$$ where $a$, $b$, and $c$ are constants. Here, $a = -2$, $b = 14$, and $c = -20$. Since $a < 0$, the parabola opens downward, meaning the profit has a maximum point. 3. **Find the vertex (maximum profit):** The vertex $b$-coordinate is found by: $$b = -\frac{B}{2A} = -\frac{14}{2 \times (-2)} = -\frac{14}{-4} = 3.5$$ 4. **Calculate the maximum profit $P$ at $b=3.5$:** $$P = -2(3.5)^2 + 14(3.5) - 20$$ $$= -2(12.25) + 49 - 20$$ $$= -24.5 + 49 - 20$$ $$= 4.5$$ 5. **Interpretation:** The maximum profit is 4.5 (in thousands) when 3.5 thousand skateboards are produced. 6. **Find the roots (break-even points):** Solve $-2b^2 + 14b - 20 = 0$. Divide both sides by $-2$: $$\cancel{-2}b^2 + \cancel{-2} \times \frac{14}{-2}b + \cancel{-2} \times \frac{-20}{-2} = 0$$ which simplifies to: $$b^2 - 7b + 10 = 0$$ Factor: $$(b - 5)(b - 2) = 0$$ So, $b = 5$ or $b = 2$. 7. **Interpretation:** The profit is zero when 2 thousand or 5 thousand skateboards are produced. --- **Final answers:** - Maximum profit: $4.5$ thousand at $b = 3.5$ thousand skateboards. - Break-even points: $b = 2$ thousand and $b = 5$ thousand skateboards.