1. **Problem 10:** The daily sales function is given by $$s = -p^2 + 120p + 1400$$ where $p$ is the price per unit. 2. **Part (a): Find the vertex by completing the square.** Start with the function: $$s = -p^2 + 120p + 1400$$ Rewrite by factoring out the negative from the quadratic terms: $$s = -(p^2 - 120p) + 1400$$ Complete the square inside the parentheses: Take half of 120, which is 60, and square it: $$60^2 = 3600$$ Add and subtract 3600 inside the parentheses: $$s = -\big(p^2 - 120p + 3600 - 3600\big) + 1400$$ $$s = -\big((p - 60)^2 - 3600\big) + 1400$$ Distribute the negative: $$s = -(p - 60)^2 + 3600 + 1400$$ $$s = -(p - 60)^2 + 5000$$ So the vertex form is: $$s = -(p - 60)^2 + 5000$$ The vertex is at $$p = 60$$ and $$s = 5000$$. 3. **Part (b): Reasonable domain and range.** - Domain: Price $p$ cannot be negative, so $$p \geq 0$$. - Also, since sales drop to zero when the price is too high or too low, find where sales are zero: Set $$s=0$$: $$0 = -(p - 60)^2 + 5000$$ $$ (p - 60)^2 = 5000$$ $$p - 60 = \pm \sqrt{5000} \approx \pm 70.71$$ So, $$p \approx 60 - 70.71 = -10.71$$ (not valid since price can't be negative) $$p \approx 60 + 70.71 = 130.71$$ Thus, domain is approximately $$0 \leq p \leq 130.71$$. - Range: Since the vertex is the maximum sales, range is $$0 \leq s \leq 5000$$. 4. **Part (c): Price for maximum sales and maximum sales.** From vertex form, maximum sales occur at $$p=60$$ with sales $$s=5000$$ units. --- 5. **Problem 11:** Height of a punted football is modeled by: $$h = -0.01x^2 + 1.18x + 2$$ where $x$ is horizontal distance and $h$ is height. 6. **Part (a): Find the vertex by completing the square.** Start with: $$h = -0.01x^2 + 1.18x + 2$$ Factor out -0.01 from the quadratic terms: $$h = -0.01(x^2 - 118x) + 2$$ Complete the square inside the parentheses: Half of 118 is 59, square is $$59^2 = 3481$$ Add and subtract 3481 inside parentheses: $$h = -0.01(x^2 - 118x + 3481 - 3481) + 2$$ $$h = -0.01((x - 59)^2 - 3481) + 2$$ Distribute -0.01: $$h = -0.01(x - 59)^2 + 0.01 \times 3481 + 2$$ $$h = -0.01(x - 59)^2 + 34.81 + 2$$ $$h = -0.01(x - 59)^2 + 36.81$$ Vertex is at $$x = 59$$ and $$h = 36.81$$ feet. 7. **Part (b): Maximum height of the punt.** Maximum height is the vertex height: $$36.81$$ feet. 8. **Part (c): Height at $x=5$ feet (to block the punt).** Calculate height at $$x=5$$: $$h = -0.01(5)^2 + 1.18(5) + 2 = -0.01(25) + 5.9 + 2 = -0.25 + 5.9 + 2 = 7.65$$ feet. The player must reach at least $$7.65$$ feet to block the punt. 9. **Part (d): Find horizontal distance when ball hits ground ($h=0$).** Set $$h=0$$: $$0 = -0.01x^2 + 1.18x + 2$$ Multiply both sides by -100 to clear decimals: $$0 = x^2 - 118x - 200$$ Use quadratic formula: $$x = \frac{118 \pm \sqrt{118^2 - 4(1)(-200)}}{2}$$ Calculate discriminant: $$118^2 = 13924$$ $$4 \times 1 \times -200 = -800$$ $$\sqrt{13924 + 800} = \sqrt{14724} \approx 121.34$$ Solutions: $$x = \frac{118 \pm 121.34}{2}$$ Positive root: $$x = \frac{118 + 121.34}{2} = \frac{239.34}{2} = 119.67$$ Negative root is negative and discarded. So, the ball hits the ground approximately $$119.67$$ feet downfield.