Quadratic Modeling

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Quadratic Modeling · Grade 10 · Worksheet 1

1. Noah throws a ball upward from a height of 8 meters with an initial velocity of 24 m/s. The height h (in meters) after t seconds is given by h(t) = -5t² + 24t + 8. Find the maximum height reached by the ball. Answer: 36.8 Solution: The height function is h(t) = -5t² + 24t + 8. Since a = -5 is negative, the parabola opens downward, so the vertex gives the maximum height. The time at the vertex is t = -b/(2a) = -24/(2*(-5)) = -24/(-10) = 2.4 seconds.
   Full step-by-step solution

   Step 1: The height function is h(t) = -5t² + 24t + 8. Since a = -5 is negative, the parabola opens downward, so the vertex gives the maximum height. Step 2: The time at the vertex is t = -b/(2a) = -24/(2*(-5)) = -24/(-10) = 2.4 seconds. Step 3: Substitute t = 2.4 into the height function: h(2.4) = -5(2.4)² + 24(2.4) + 8. Step 4: Calculate (2.4)² = 5.76. Then -5(5.76) = -28.8. Also 24(2.4) = 57.6. Step 5: So h(2.4) = -28.8 + 57.6 + 8 = 36.8. Step 6: The maximum height reached by the ball is 36.8 meters.

2. Mason throws a ball upward from a height of 2 meters with an initial velocity of 27 m/s. The height h (in meters) after t seconds is given by h(t) = -5t² + 27t + 2. Find the maximum height reached by the ball. Answer: 38.45 Solution: Identify the coefficients: a = -5, b = 27, c = 2. The time at the vertex is t = -b/(2a) = -27/(2 * -5) = -27/(-10) = 2.7 seconds. Substitute t = 2.7 into the height equation: h(2.7) = -5(2.7)² + 27(2.7) + 2.
   Full step-by-step solution

   Step 1: Identify the coefficients: a = -5, b = 27, c = 2. Step 2: The time at the vertex is t = -b/(2a) = -27/(2 * -5) = -27/(-10) = 2.7 seconds. Step 3: Substitute t = 2.7 into the height equation: h(2.7) = -5(2.7)² + 27(2.7) + 2. Step 4: Calculate (2.7)² = 7.29. Step 5: -5(7.29) = -36.45. Step 6: 27(2.7) = 72.9. Step 7: h(2.7) = -36.45 + 72.9 + 2 = 38.45. The maximum height reached by the ball is 38.45 meters.

3. Mason is a farmer who wants to build a rectangular enclosure for his sheep against a straight riverbank. He has 72 meters of fencing. The side along the river does not need fencing. The width of the enclosure (the sides perpendicular to the river) is x meters, and the length (parallel to the river) is y meters. Mason wants to maximize the area of the enclosure. What width x (in meters) will give the maximum area? Answer: 18 Solution: The fencing is used for two widths and one length: 2x + y = 72. So y = 72 - 2x. The area A = x * y = x(72 - 2x) = 72x - 2x^2.
   Full step-by-step solution

   Step 1: The fencing is used for two widths and one length: 2x + y = 72. So y = 72 - 2x. Step 2: The area A = x * y = x(72 - 2x) = 72x - 2x^2. Step 3: This is a quadratic function A(x) = -2x^2 + 72x. The coefficient of x^2 is negative, so the parabola opens downward and the vertex gives the maximum. Step 4: For a quadratic ax^2 + bx + c, the vertex occurs at x = -b/(2a). Here a = -2, b = 72, so x = -72/(2 * -2) = -72/(-4) = 18. Step 5: Therefore, the width that maximizes the area is 18 meters. The answer is 18.

4. A parabolic arch bridge spans a river with its vertex at the highest point of the arch. The arch is modeled by the quadratic function h(x) = -0.02x² + 24, where h(x) represents the height in meters above the water level and x represents the horizontal distance in meters from the center of the arch. The bridge's supports are located where the arch meets the water (h(x) = 0). What is the total span of the bridge between these two support points? Answer: 69.28 Solution: h(x) = -0.02x² + 24 The bridge supports are where the arch meets the water, so h(x) = 0.
   Full step-by-step solution

   We are given the function for the arch: h(x) = -0.02x² + 24 The bridge supports are where the arch meets the water, so h(x) = 0. Step 1: Set h(x) = 0 -0.02x² + 24 = 0 Step 2: Solve for x² -0.02x² = -24 Multiply both sides by -1: 0.02x² = 24 Step 3: Divide both sides by 0.02 x² = 24 / 0.02 x² = 1200 Step 4: Take the square root x = ±√1200 Simplify √1200: √1200 = √(400 × 3) = 20√3 So x = 20√3 and x = -20√3 Step 5: Interpret the result These are the horizontal distances from the center of the arch to each support. The total span is the distance between these two points: Total span = (20√3) - (-20√3) = 40√3 Step 6: Compute numerical value √3 ≈ 1.732 40 × 1.732 = 69.28 Step 7: Conclusion The total span of the bridge is 69.28 meters.

5. Liam is designing a rectangular garden next to a straight stone wall. He has 120 meters of fencing to enclose the other three sides of the garden (the side along the wall does not need fencing). Liam wants to maximize the area of the garden. What dimensions (length parallel to the wall and width perpendicular to the wall) will give the maximum area, and what is that maximum area? Answer: Length = 60 m, Width = 30 m, Maximum area = 1800 m² Solution: Let x be the width perpendicular to the wall (in meters) and y be the length parallel to the wall (in meters). The fencing is used for two widths and one length: 2x + y = 120. Solve for y: y = 120 - 2x.
   Full step-by-step solution

   Step 1: Let x be the width perpendicular to the wall (in meters) and y be the length parallel to the wall (in meters). The fencing is used for two widths and one length: 2x + y = 120. Solve for y: y = 120 - 2x. Step 2: The area A = x * y = x(120 - 2x) = 120x - 2x². Step 3: This is a quadratic in the form A = -2x² + 120x. Since the coefficient of x² is negative, the parabola opens downward and the vertex gives the maximum area. Step 4: The x-coordinate of the vertex is x = -b/(2a) where a = -2, b = 120. So x = -120/(2 * -2) = -120/-4 = 30. Step 5: Substitute x = 30 into y = 120 - 2(30) = 120 - 60 = 60. Step 6: The maximum area is A = 30 * 60 = 1800 m². The dimensions that maximize the area are: length (parallel to wall) = 60 m, width (perpendicular to wall) = 30 m, and the maximum area is 1800 m².

6. The profit P (in dollars) from selling x units of a product is modeled by P(x) = -2x² + 120x - 1000. Find the number of units that maximizes profit. Answer: 30 Solution: The profit function is P(x) = -2x² + 120x - 1000 Since the coefficient of x² is negative (-2), the parabola opens downward, so the vertex gives the maximum profit Use the vertex formula x = -b/(2a) where a = -2 and b = 120 x = -120/(2 × -2) = -120/(-4) = 30 Therefore, selling 30 units maximizes…
   Full step-by-step solution

   Step 1: The profit function is P(x) = -2x² + 120x - 1000 Step 2: Since the coefficient of x² is negative (-2), the parabola opens downward, so the vertex gives the maximum profit Step 3: Use the vertex formula x = -b/(2a) where a = -2 and b = 120 Step 4: x = -120/(2 × -2) = -120/(-4) = 30 Step 5: Therefore, selling 30 units maximizes the profit The answer is 30.

7. Olivia is designing a rectangular garden for a community project. The garden will be enclosed on three sides by a fence (the fourth side is against a wall of a building). She has 40 meters of fencing available. Let x represent the width of the garden (the sides perpendicular to the wall). Write a quadratic function A(x) that models the area of the garden in square meters, and determine the width x that maximizes the area. What is the width that gives the maximum area? Answer: 10 Solution: Let x be the width of the garden (distance from the wall). The fencing covers the two widths and one length (since the fourth side is the wall). So, 2x + length = 40.
   Full step-by-step solution

   Step 1: Let x be the width of the garden (distance from the wall). The fencing covers the two widths and one length (since the fourth side is the wall). So, 2x + length = 40. Therefore, length = 40 - 2x. Step 2: The area A(x) = width * length = x(40 - 2x) = 40x - 2x^2. Step 3: Write it in standard form: A(x) = -2x^2 + 40x. Step 4: For a quadratic in the form ax^2 + bx + c, the vertex (maximum point) occurs at x = -b/(2a). Here, a = -2, b = 40. Step 5: x = -40 / (2 * -2) = -40 / -4 = 10. Step 6: The width that maximizes the area is 10 meters. The answer is 10.