Quadratic Modeling

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Quadratic Modeling · Grade 10 · Worksheet 2

1. 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.

2. Matiu is designing a water fountain for a public park. The water stream from the fountain follows a parabolic path. The height of the water (in meters) above the nozzle is modeled by the quadratic function h(x) = -0.02x² + 0.8x + 1.2, where x is the horizontal distance from the nozzle in meters. Matiu wants to know the maximum height the water reaches and the horizontal distance at which this maximum height occurs. What is the maximum height of the water stream? Answer: 9.2 Solution: The quadratic function is h(x) = -0.02x² + 0.8x + 1.2. This is in standard form ax² + bx + c, where a = -0.02, b = 0.8, and c = 1.2.
   Full step-by-step solution

   Step 1: The quadratic function is h(x) = -0.02x² + 0.8x + 1.2. This is in standard form ax² + bx + c, where a = -0.02, b = 0.8, and c = 1.2. Step 2: Since a is negative, the parabola opens downward, and the vertex gives the maximum height. Step 3: The x-coordinate of the vertex is found using the formula x = -b/(2a). Step 4: Substitute b = 0.8 and a = -0.02: x = -(0.8) / (2 * -0.02) = -0.8 / -0.04 = 20. Step 5: The maximum height occurs at a horizontal distance of 20 meters from the nozzle. Step 6: To find the maximum height, substitute x = 20 into the function: h(20) = -0.02(20)² + 0.8(20) + 1.2. Step 7: Calculate: -0.02(400) + 16 + 1.2 = -8 + 16 + 1.2 = 9.2. Step 8: The maximum height of the water stream is 9.2 meters.

3. 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.

4. A rectangular garden is to be enclosed with 100 meters of fencing, with one side against a building (requiring no fence). If the area A of the garden is given by A(x) = x(100 - 2x), where x is the length of the side perpendicular to the building, find the maximum possible area. Answer: 1250 Solution: The area function is A(x) = x(100 - 2x) = -2x² + 100x. This is a quadratic function opening downward (a = -2 < 0), so it has a maximum at its vertex.
   Full step-by-step solution

   Step 1: The area function is A(x) = x(100 - 2x) = -2x² + 100x. Step 2: This is a quadratic function opening downward (a = -2 < 0), so it has a maximum at its vertex. Step 3: The x-coordinate of the vertex is given by x = -b/(2a) = -100/(2*(-2)) = -100/(-4) = 25. Step 4: Substitute x = 25 into the area function: A(25) = -2(25)² + 100(25) = -2(625) + 2500 = -1250 + 2500 = 1250. Step 5: The maximum possible area is 1250 square meters.

5. Matiu launches a model rocket from a platform 12 meters above the ground. The rocket's height h(t) in meters after t seconds is given by h(t) = -4.9t² + 49t + 12. Find the maximum height reached by the rocket. Answer: 134.5 Solution: The height function is h(t) = -4.9t² + 49t + 12. Since a = -4.9 < 0, the parabola opens downward, so the vertex gives the maximum height. The time at the vertex is given by t = -b/(2a).
   Full step-by-step solution

   Step 1: The height function is h(t) = -4.9t² + 49t + 12. Since a = -4.9 < 0, the parabola opens downward, so the vertex gives the maximum height. Step 2: The time at the vertex is given by t = -b/(2a). Here, a = -4.9 and b = 49. Step 3: t = -49 / (2 * -4.9) = -49 / (-9.8) = 5 seconds. Step 4: Substitute t = 5 into the height function: h(5) = -4.9(5)² + 49(5) + 12 = -4.9(25) + 245 + 12 = -122.5 + 245 + 12 = 134.5. Step 5: The maximum height reached by the rocket is 134.5 meters.

6. A rectangular field is to be fenced on three sides, with the fourth side against a river. Tane has 120 meters of fencing. The area A of the field, as a function of the width x (in meters) perpendicular to the river, is A(x) = x(120 - 2x). Find the maximum possible area of the field. Answer: 1800 Solution: Expand the area function: A(x) = x(120 - 2x) = -2x² + 120x. This is a quadratic function with a = -2 and b = 120. Since a < 0, the parabola opens downward, so the vertex gives the maximum area.
   Full step-by-step solution

   Step 1: Expand the area function: A(x) = x(120 - 2x) = -2x² + 120x. Step 2: This is a quadratic function with a = -2 and b = 120. Since a < 0, the parabola opens downward, so the vertex gives the maximum area. Step 3: The x-coordinate of the vertex is given by x = -b/(2a). Substitute a = -2 and b = 120: x = -120/(2 * -2) = -120/(-4) = 30. Step 4: Substitute x = 30 into the area function: A(30) = -2(30)² + 120(30) = -2(900) + 3600 = -1800 + 3600 = 1800. Step 5: The maximum possible area is 1800 square meters.

7. Mere launches a model rocket from a platform. The height h (in meters) of the rocket t seconds after launch is given by h(t) = -4t² + 24t + 28. Find the maximum height reached by the rocket. Answer: 64 Solution: The height function is h(t) = -4t² + 24t + 28. This is a quadratic function with a = -4, b = 24, and c = 28. Since a = -4 is negative, the parabola opens downward, so the vertex represents the maximum height.
   Full step-by-step solution

   Step 1: The height function is h(t) = -4t² + 24t + 28. This is a quadratic function with a = -4, b = 24, and c = 28. Step 2: Since a = -4 is negative, the parabola opens downward, so the vertex represents the maximum height. Step 3: The time at the vertex is given by t = -b/(2a) = -24/(2 * -4) = -24/(-8) = 3 seconds. Step 4: Substitute t = 3 into the height function: h(3) = -4(3)² + 24(3) + 28 = -4(9) + 72 + 28 = -36 + 72 + 28 = 64. Step 5: The maximum height reached by the rocket is 64 meters.