Quadratic Contextual Problems - Math 2

1. A soccer ball is kicked from ground level with an initial velocity such that its height h (in feet) at time t seconds is h(t) = -16t² + 48t. A player wants to head the ball when it's 32 feet high. At what times will the ball be at this height?
   Answer: At 1 second and 2 seconds, during ascent and descent phases respectively When you encounter projectile motion problems, you're dealing with quadratic equations where an object follows a parabolic path. The key insight is that most projectiles reach any given height twice—once while going up and once while coming down. To find when the ball is at 32 feet, set the height equation equal to 32 and solve: $-16t^2 + 48t = 32$ Rearranging: $-16t^2 + 48t - 32 = 0$ Dividing by -16: $t^2 - 3t + 2 = 0$ This factors as $(t - 1)(t - 2) = 0$, giving us $t = 1$ second and $t = 2$ seconds. The ball reaches 32 feet at 1 second (ascending) and 2 seconds (descending), confirming answer D. Looking at the wrong answers: A claims the ball only reaches 32 feet once at maximum height, but the maximum occurs at $t = 1.5$ seconds with height $h(1.5) = 36$ feet—not 32 feet. B gives times 0.8 and 2.2 seconds, which you can verify don't satisfy the equation when substituted back. C offers 1.2 and 1.8 seconds, which also fail when checked against the original equation. Remember that projectile motion problems typically have two solutions for any height below the maximum. Always factor or use the quadratic formula to find both times, and verify your answers by substituting back into the original equation. This double-checking catches calculation errors and confirms you haven't missed a solution.
2. A stone is thrown horizontally from a cliff 144 feet high. The stone's height above the ground after $t$ seconds is $h(t) = 144 - 16t^2$. At the same time, a second stone is thrown upward from the ground with height $h_2(t) = -16t^2 + 48t$. When do the stones hit the ground, and which stone is higher at $t = 1.5$ seconds?
   Answer: First stone hits at t = 3 seconds, second at t = 3 seconds; first stone is higher at t = 1.5 For the first stone: 144 - 16t² = 0, so 16t² = 144, giving t² = 9, so t = 3 seconds. For the second stone: -16t² + 48t = 0, so t(-16t + 48) = 0, giving t = 0 or t = 3 seconds. Both stones hit the ground at t = 3 seconds. At t = 1.5: h₁(1.5) = 144 - 16(2.25) = 144 - 36 = 108 feet. h₂(1.5) = -16(2.25) + 48(1.5) = -36 + 72 = 36 feet. The first stone is higher at t = 1.5 seconds.
3. A rocket is launched from a 48-foot platform. Its height h (in feet) above ground at time t seconds is h(t) = -16t² + 80t + 48. The rocket will deploy a parachute when it returns to the same height as the launch platform during its descent. When will this occur?
   Answer: 5 seconds after launch, when the rocket returns to platform height When you encounter projectile motion problems, you're working with quadratic functions where the object follows a parabolic path up and then back down. The key insight here is understanding what "returns to platform height" means mathematically. The rocket starts at 48 feet (the platform height), so you need to find when $h(t) = 48$ again during its descent. Setting up the equation: $-16t^2 + 80t + 48 = 48$ Subtracting 48 from both sides: $-16t^2 + 80t = 0$ Factoring out the common terms: $-16t(t - 5) = 0$ This gives you $t = 0$ or $t = 5$. Since $t = 0$ represents the launch moment, $t = 5$ seconds is when the rocket returns to platform height during descent. Looking at the wrong answers: A) 4.8 seconds might result from calculation errors in the quadratic formula or decimal approximation mistakes. C) 4.5 seconds could come from incorrectly setting up the equation or arithmetic errors during factoring. D) 5.2 seconds might arise from using the quadratic formula incorrectly or making computational mistakes. The correct answer is B) 5 seconds after launch. Study tip: In projectile problems, always check if you can factor before using the quadratic formula. Many exam problems are designed with "nice" numbers that factor cleanly. Also, remember that there are typically two solutions—one for the way up and one for the way down—so identify which time corresponds to the phase of motion the question asks about.
4. A manufacturer finds that the daily profit $P$ (in dollars) from producing $x$ items is $P(x) = -2x^2 + 160x - 1800$. Due to equipment limitations, they can produce at most 60 items per day. What is the optimal daily production level, and what happens if they must produce at least 50 items due to contractual obligations?
   Answer: Optimal production is 40 items with profit $1,400; producing 50 items gives $1,200 profit which is still viable The vertex occurs at x = -160/(-4) = 40 items. P(40) = -2(1600) + 160(40) - 1800 = -3200 + 6400 - 1800 = 1400 dollars. At x = 50: P(50) = -2(2500) + 160(50) - 1800 = -5000 + 8000 - 1800 = 1200 dollars. At x = 60: P(60) = -2(3600) + 160(60) - 1800 = -7200 + 9600 - 1800 = 600 dollars. The optimal production is 40 items ($1,400 profit), but if contractually obligated to produce 50 items, the profit drops to $1,200, which is still substantial and viable.
5. The company policy states that if daily profit falls below $2,100, overtime must be reduced. For what ranges of overtime hours will the company need to implement this policy?
   Answer: When overtime exceeds 7 hours per day or falls below 1 hour per day When you encounter a quadratic function modeling real-world scenarios like profit, you're typically looking for specific values or ranges where the function meets certain conditions. Here, you need to find when profit falls below $2,100. Since profit P is given in hundreds of dollars, $2,100 equals 21 hundreds. You need to solve the inequality $P(h) < 21$, which means $-h^2 + 8h + 20 < 21$. Rearranging: $-h^2 + 8h + 20 - 21 < 0$, so $-h^2 + 8h - 1 < 0$, or equivalently $h^2 - 8h + 1 > 0$. Using the quadratic formula to find where $h^2 - 8h + 1 = 0$: $h = \frac{8 \pm \sqrt{64 - 4}}{2} = \frac{8 \pm \sqrt{60}}{2} = \frac{8 \pm 2\sqrt{15}}{2} = 4 \pm \sqrt{15}$ Since $\sqrt{15} \approx 3.87$, the roots are approximately $h \approx 0.13$ and $h \approx 7.87$. Since the coefficient of $h^2$ is positive, the parabola opens upward, making the inequality true when $h < 0.13$ or $h > 7.87$. Choice B correctly identifies this as "below 1 hour" and "exceeds 7 hours" (using reasonable rounding). Choice A uses incorrect values of -1 and 9 hours. Choice C gives 2 and 6 hours, which are too far from the calculated boundaries. Choice D is wrong because the policy clearly needs implementation at extreme overtime values. Remember: when solving quadratic inequalities, always find the roots first, then determine which intervals satisfy your inequality based on the parabola's direction.
6. The company will break even (zero profit) when its revenue equals its fixed costs of $15,000. For what production levels will the company break even?
   Answer: 500 units and 700 units, representing the two break-even production levels When you encounter break-even problems, you're looking for the point where revenue equals costs. This requires setting up and solving a quadratic equation. To find the break-even points, set the revenue function equal to the fixed costs. Since R(x) is in thousands of dollars and the fixed costs are $15,000, you need: $-x^2 + 12x - 20 = 15$ Rearranging to standard form: $-x^2 + 12x - 35 = 0$ Multiplying by -1: $x^2 - 12x + 35 = 0$ Using the quadratic formula or factoring: $(x - 5)(x - 7) = 0$ This gives x = 5 and x = 7. Since x represents hundreds of units, the break-even points are 500 units and 700 units. Choice A (300 and 900 units) results from incorrectly solving the equation or misinterpreting the variable scaling. Choice B (350 and 850 units) likely comes from calculation errors in the quadratic formula. Choice C (400 and 800 units) might result from setting up the equation incorrectly or making arithmetic mistakes during factoring. The correct answer is D: 500 units and 700 units represent the two break-even production levels. For quadratic business problems, always pay careful attention to the units given in the problem. Here, x represented hundreds of units, so your final answer needed to be converted accordingly. Also, remember that break-even problems typically yield two solutions because revenue functions are parabolic—there's usually a lower and upper production level where costs equal revenue.
7. A photographer wants to capture the projectile when it's at 75% of its maximum height during its descent phase. At what time should the photograph be taken?
   Answer: 4.25 seconds after launch, when the projectile is descending First, find the maximum height. The vertex occurs at t = 29.4/(2×4.9) = 3 seconds. Maximum height = -4.9(3)² + 29.4(3) = 44.1 meters. 75% of maximum = 0.75 × 44.1 = 33.075 meters. Solve -4.9t² + 29.4t = 33.075. Rearranging: -4.9t² + 29.4t - 33.075 = 0. Using the quadratic formula: t = (29.4 ± √(29.4² - 4(-4.9)(-33.075)))/(2(-4.9)). This gives t ≈ 1.75 seconds (ascending) and t ≈ 4.25 seconds (descending). Since we want the descent phase, t = 4.25 seconds.
8. The profit P (in thousands of dollars) for a small business is modeled by P(x) = -2x² + 24x - 40, where x is the number of months after opening. During which time periods will the business have a profit of at least $30,000?
   Answer: Between months 5 and 7, including both endpoints of this interval We need P(x) ≥ 30, so -2x² + 24x - 40 ≥ 30. This gives -2x² + 24x - 70 ≥ 0, or 2x² - 24x + 70 ≤ 0, or x² - 12x + 35 ≤ 0. Factoring: (x - 5)(x - 7) ≤ 0. This inequality is satisfied when 5 ≤ x ≤ 7. We can verify: at x = 5, P(5) = -2(25) + 24(5) - 40 = 30; at x = 7, P(7) = -2(49) + 24(7) - 40 = 30. Both endpoints give exactly $30,000 profit, so they're included.
9. A stone is thrown horizontally from a cliff 144 feet high. Its height h (in feet) above the ground t seconds after being thrown is h(t) = 144 - 16t². A rescue team is positioned 60 feet above ground level. During what time interval could they potentially catch the stone?
   Answer: The rescue team cannot catch the stone as it passes their level too quickly The stone reaches the 60-foot level when 144 - 16t² = 60, so 16t² = 84, giving t² = 5.25, so t = √5.25 ≈ 2.29 seconds. However, this is just one instant in time, not an interval. Since the stone is in free fall (not thrown upward), it passes through each height exactly once, so there's no time interval during which it remains at or near the 60-foot level. The rescue team would need to catch it at precisely t ≈ 2.29 seconds, which is practically impossible.
10. A farmer has 200 feet of fencing to create a rectangular pen against an existing barn wall. The barn wall will serve as one side of the rectangle, so fencing is only needed for three sides. What dimensions maximize the area, and what is this maximum area?
    Answer: 50 feet by 100 feet rectangle with maximum area of 5,000 square feet Let x = width perpendicular to barn, y = length parallel to barn. We need 2x + y = 200, so y = 200 - 2x. Area = xy = x(200 - 2x) = 200x - 2x². To maximize, take derivative: dA/dx = 200 - 4x = 0, giving x = 50. Then y = 200 - 2(50) = 100. Maximum area = 50 × 100 = 5,000 square feet. We can verify this is a maximum since d²A/dx² = -4 < 0.
11. If the company wants to install a decorative ring at a height where the water passes through exactly twice during each water cycle, what heights are possible for this ring placement?
    Answer: Any height between 4 feet and 40 feet, excluding the maximum height of 40 feet For the water to pass through a ring exactly twice, the equation h(t) = k must have exactly 2 solutions, where k is the ring height. The vertex form shows the maximum height occurs at t = 48/(2×16) = 1.5 seconds, giving h(1.5) = -16(1.5)² + 48(1.5) + 4 = 40 feet. At t = 0, h = 4 feet (initial height). For exactly 2 solutions, k must be between the initial height (4 feet) and maximum height (40 feet), but not equal to the maximum where there would be only 1 solution.
12. The number of bacteria in a culture is modeled by $N(t) = 1000 + 200t - 5t^2$, where $t$ is time in hours and $N(t)$ is the number of bacteria. After how many hours will the bacteria population return to its initial value, and what does this suggest about the long-term viability of the culture?
    Answer: After 40 hours; the culture will continue to decline toward extinction The initial population is N(0) = 1000. To find when it returns to this value: 1000 + 200t - 5t² = 1000. This simplifies to 200t - 5t² = 0, or t(200 - 5t) = 0. So t = 0 or t = 40. The population returns to 1000 after 40 hours. Since the coefficient of t² is negative (-5), this is a downward-opening parabola. After reaching its maximum (at t = -200/(-10) = 20 hours), the population declines. After t = 40, the population will be less than the initial value and continue decreasing, suggesting the culture is not sustainable long-term.
13. A bridge's arch can be modeled by $h(x) = -0.1x^2 + 4x$, where $h$ is the height in feet and $x$ is the horizontal distance in feet from one end. A boat with a height of 35 feet needs to pass under the bridge. What is the range of horizontal positions where the boat can safely pass?
    Answer: Between 10 and 30 feet from the starting end Set h(x) = 35: -0.1x² + 4x = 35. Rearranging: -0.1x² + 4x - 35 = 0, or x² - 40x + 350 = 0. Using the quadratic formula: x = (40 ± √(1600 - 1400))/2 = (40 ± √200)/2 = (40 ± 10√2)/2 = 20 ± 5√2. Since √2 ≈ 1.414, we get x ≈ 20 ± 7.07, so x ≈ 12.93 or x ≈ 27.07. The boat can safely pass between approximately 13 and 27 feet. Option B (10 to 30 feet) provides the safe range with appropriate safety margins.

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