1. **Problem 1: Profit from assembling bicycles** Given: $$P = -x^2 + 50x - 300$$ where $P$ is profit and $x$ is number of bicycles. **a) Loss if no bicycles are assembled:** Substitute $x=0$: $$P = -0^2 + 50\times0 - 300 = -300$$ Loss is 300. **b) Number of bicycles to maximize profit:** The profit function is a quadratic with $a = -1 < 0$, so it opens downward and has a maximum at vertex. Vertex formula for $x$ is: $$x = -\frac{b}{2a} = -\frac{50}{2 \times (-1)} = \frac{50}{2} = 25$$ **c) Maximum profit:** Substitute $x=25$ into $P$: $$P = -(25)^2 + 50 \times 25 - 300 = -625 + 1250 - 300 = 325$$ 2. **Problem 2: Cost per lamp** Given: $$C = x^2 - 16x + 67$$ where $C$ is cost per lamp and $x$ is number of lamps. **a) Number of lamps to minimize cost:** Since $a=1 > 0$, parabola opens upward and minimum is at vertex: $$x = -\frac{b}{2a} = -\frac{-16}{2 \times 1} = \frac{16}{2} = 8$$ **b) Minimum cost per lamp:** Substitute $x=8$: $$C = 8^2 - 16 \times 8 + 67 = 64 - 128 + 67 = 3$$ **c) Cost per lamp if 5 lamps made:** Substitute $x=5$: $$C = 5^2 - 16 \times 5 + 67 = 25 - 80 + 67 = 12$$ 3. **Problem 3: Hourly profit from taxis** Given: $$P = -2n^2 + 120n - 200$$ where $P$ is profit and $n$ is number of taxis. **a) Number of taxis for maximum profit:** Since $a = -2 < 0$, maximum at vertex: $$n = -\frac{b}{2a} = -\frac{120}{2 \times (-2)} = \frac{120}{4} = 30$$ **b) Maximum hourly profit:** Substitute $n=30$: $$P = -2 \times 30^2 + 120 \times 30 - 200 = -2 \times 900 + 3600 - 200 = -1800 + 3600 - 200 = 1600$$ **c) Hourly loss if no taxis:** Substitute $n=0$: $$P = -2 \times 0 + 120 \times 0 - 200 = -200$$ Loss is 200.