1. **Problem Statement:** John is selling ski packages at a price of 1400 with 16 expected buyers. For every 50 decrease in price, 2 more students buy the package. We need to find: a) Axis of symmetry b) Coordinates of the vertex c) Maximum revenue d) Ticket price that maximizes revenue e) Number of 50 decreases that produce no revenue 2. **Define variables:** Let $x$ be the number of $50 decreases in price. Price per package: $p = 1400 - 50x$ Number of buyers: $n = 16 + 2x$ 3. **Revenue function:** Revenue $R = p \times n = (1400 - 50x)(16 + 2x)$ 4. **Expand revenue:** $$R = 1400 \times 16 + 1400 \times 2x - 50x \times 16 - 50x \times 2x$$ $$R = 22400 + 2800x - 800x - 100x^2$$ $$R = -100x^2 + 2000x + 22400$$ 5. **Rewrite revenue function:** $$R(x) = -100x^2 + 2000x + 22400$$ 6. **a) Axis of symmetry:** For quadratic $ax^2 + bx + c$, axis is $x = -\frac{b}{2a}$ Here, $a = -100$, $b = 2000$ $$x = -\frac{2000}{2 \times -100} = -\frac{2000}{-200} = 10$$ 7. **b) Coordinates of vertex:** Vertex $x$-coordinate is 10. Calculate $R(10)$: $$R(10) = -100(10)^2 + 2000(10) + 22400 = -100(100) + 20000 + 22400 = -10000 + 20000 + 22400 = 32400$$ Vertex coordinates: $(10, 32400)$ 8. **c) Maximum revenue:** Maximum revenue is the vertex's $y$-value: $32400$ 9. **d) Ticket price that maximizes revenue:** Price at $x=10$ decreases: $$p = 1400 - 50 \times 10 = 1400 - 500 = 900$$ 10. **e) Number of $50 decreases that produce no revenue:** Set revenue to zero: $$0 = -100x^2 + 2000x + 22400$$ Divide both sides by -100: $$0 = \cancel{-100}x^2 + \cancel{-100} \times -20x + \cancel{-100} \times -224$$ $$0 = x^2 - 20x - 224$$ Solve quadratic: $$x = \frac{20 \pm \sqrt{(-20)^2 - 4(1)(-224)}}{2} = \frac{20 \pm \sqrt{400 + 896}}{2} = \frac{20 \pm \sqrt{1296}}{2} = \frac{20 \pm 36}{2}$$ Two solutions: $$x_1 = \frac{20 + 36}{2} = 28$$ $$x_2 = \frac{20 - 36}{2} = -8$$ (discard negative) Number of $50 decreases producing no revenue is $28$. --- **Final answers:** a) Axis of symmetry: $x=10$ b) Vertex: $(10, 32400)$ c) Maximum revenue: $32400$ d) Ticket price maximizing revenue: $900$ e) Number of $50 decreases producing no revenue: $28$