Quadratic Modeling

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Quadratic Modeling · Grade 10 · Worksheet 3

1. A company's profit P(x) = -2x² + 80x - 600, where x is the number of units sold. Find the number of units that maximizes profit. Answer: 20 Solution: The profit function is P(x) = -2x² + 80x - 600. Since the coefficient of x² is negative (-2), the parabola opens downward, and the vertex represents the maximum point.
   Full step-by-step solution

   Step 1: The profit function is P(x) = -2x² + 80x - 600. Since the coefficient of x² is negative (-2), the parabola opens downward, and the vertex represents the maximum point. Step 2: For a quadratic function in the form ax² + bx + c, the x-coordinate of the vertex is given by x = -b/(2a). Step 3: Here, a = -2 and b = 80. Step 4: Substitute into the formula: x = -80/(2 * -2) = -80/(-4) = 20. Step 5: Therefore, the number of units that maximizes profit is 20.

2. A rectangular garden has a perimeter of 60 meters. If the area is modeled by A(x) = x(30 - x), find the maximum possible area of the garden. Answer: 225 Solution: The area function is A(x) = x(30 - x) = -x² + 30x This is a quadratic function in the form ax² + bx + c, where a = -1, b = 30, c = 0 Since a < 0, the parabola opens downward, so the vertex gives the maximum value The x-coordinate of the vertex is given by x = -b/(2a) = -30/(2×-1) = -30/-2 = 15…
   Full step-by-step solution

   Step 1: The area function is A(x) = x(30 - x) = -x² + 30x Step 2: This is a quadratic function in the form ax² + bx + c, where a = -1, b = 30, c = 0 Step 3: Since a < 0, the parabola opens downward, so the vertex gives the maximum value Step 4: The x-coordinate of the vertex is given by x = -b/(2a) = -30/(2×-1) = -30/-2 = 15 Step 5: Substitute x = 15 into the area function: A(15) = 15(30 - 15) = 15 × 15 = 225 Step 6: The maximum possible area is 225 square meters.

3. Ava is designing a parabolic arch for a new garden entrance. The arch will be made of steel and its height (in meters) above the ground at a horizontal distance x (in meters) from the left base is modeled by the quadratic function h(x) = -0.25x² + 3x + 1.5. The arch is supported by two vertical pillars at the left and right bases where the arch meets the ground. Determine the maximum height of the arch and the horizontal distance from the left base where this maximum height occurs. Answer: Maximum height of 10.5 meters at a horizontal distance of 6 meters from the left base Solution: The quadratic function is h(x) = -0.25x² + 3x + 1.5, where a = -0.25, b = 3, c = 1.5. Since a is negative, the parabola opens downward, so the vertex gives the maximum height.
   Full step-by-step solution

   Step 1: The quadratic function is h(x) = -0.25x² + 3x + 1.5, where a = -0.25, b = 3, c = 1.5. Since a is negative, the parabola opens downward, so the vertex gives the maximum height. Step 2: The x-coordinate of the vertex for a quadratic in the form ax² + bx + c is given by x = -b/(2a). Step 3: Substitute the values: x = -3/(2 * -0.25) = -3/(-0.5) = 6. Step 4: The maximum height occurs at a horizontal distance of 6 meters from the left base. Step 5: To find the maximum height, substitute x = 6 into the function: h(6) = -0.25(6)² + 3(6) + 1.5 = -0.25(36) + 18 + 1.5 = -9 + 18 + 1.5 = 10.5. Step 6: The maximum height of the arch is 10.5 meters. The answer is: Maximum height of 10.5 meters at a horizontal distance of 6 meters from the left base.

4. Emma throws a ball upward from a height of 3 meters with an initial velocity of 21 m/s. The height h (in meters) after t seconds is given by h(t) = -5t² + 21t + 3. Find the maximum height reached by the ball. Answer: 25.05 Solution: The height function is h(t) = -5t² + 21t + 3. This is a quadratic function with a = -5, b = 21, and c = 3. Since a = -5 < 0, the parabola opens downward, so the vertex gives the maximum height.
   Full step-by-step solution

   Step 1: The height function is h(t) = -5t² + 21t + 3. This is a quadratic function with a = -5, b = 21, and c = 3. Since a = -5 < 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). Substitute a = -5 and b = 21: t = -21/(2 * -5) = -21/(-10) = 2.1 seconds. Step 3: Substitute t = 2.1 into the height function to find the maximum height: h(2.1) = -5(2.1)² + 21(2.1) + 3. Step 4: Calculate (2.1)² = 4.41. Then -5(4.41) = -22.05. Next, 21(2.1) = 44.1. So h(2.1) = -22.05 + 44.1 + 3 = 25.05. Step 5: The maximum height reached by the ball is 25.05 meters.

5. Liam is designing a rectangular garden in his backyard. He has 80 meters of fencing to enclose the garden. One side of the garden will be against the house, so no fencing is needed there. Liam wants to maximize the area of the garden. Let x represent the length (in meters) of each of the two sides perpendicular to the house. Write a quadratic function for the area A(x) in terms of x, then determine the dimensions that give the maximum area and the maximum area itself. Answer: x = 20 m, length parallel to house = 40 m, maximum area = 800 square meters Solution: Let x be the length of each side perpendicular to the house (in meters). The side parallel to the house has length y. Total fencing is 80 m, but only three sides need fencing: two of length x and one of length y.
   Full step-by-step solution

   Step 1: Let x be the length of each side perpendicular to the house (in meters). The side parallel to the house has length y. Total fencing is 80 m, but only three sides need fencing: two of length x and one of length y. So 2x + y = 80. Solve for y: y = 80 - 2x. Step 2: Area A = x * y = x(80 - 2x) = 80x - 2x^2. This is a quadratic function A(x) = -2x^2 + 80x. Step 3: For a quadratic in form ax^2 + bx + c, the vertex (maximum if a < 0) occurs at x = -b/(2a). Here a = -2, b = 80, so x = -80/(2 * -2) = -80/-4 = 20. Step 4: Substitute x = 20 into y = 80 - 2x: y = 80 - 40 = 40. So the dimensions are x = 20 m (perpendicular sides) and y = 40 m (side parallel to house). Step 5: Maximum area A = 20 * 40 = 800 square meters. Or substitute into A(x): A(20) = -2(20)^2 + 80(20) = -800 + 1600 = 800. The answer is x = 20 m, length parallel to house = 40 m, maximum area = 800 square meters.

6. A rectangular garden has a perimeter of 60 meters. If the area is maximized, what is the length of one side? Answer: 15 Solution: Let the length be L and width be W. The perimeter is 2L + 2W = 60. Simplify to L + W = 30, so W = 30 - L.
   Full step-by-step solution

   Step 1: Let the length be L and width be W. The perimeter is 2L + 2W = 60. Step 2: Simplify to L + W = 30, so W = 30 - L. Step 3: The area A = L × W = L(30 - L) = 30L - L². Step 4: This is a quadratic function A = -L² + 30L, which opens downward. Step 5: The maximum occurs at the vertex L = -b/(2a) = -30/(2×-1) = 15. Step 6: When L = 15, W = 30 - 15 = 15, so it's a square. The length of one side is 15 meters.

7. Mere is designing a rectangular garden next to her house. She only needs to fence three sides because the house will serve as the fourth side. She has 48 meters of fencing material. What dimensions (length and width) will maximize the area of the garden, and what is the maximum area? Answer: Length = 24 m, Width = 12 m, Maximum area = 288 m² Solution: Let the width be w meters (the sides perpendicular to the house) and the length be L meters (the side parallel to the house). Since only three sides are fenced, the total fencing is L + 2w = 48.
   Full step-by-step solution

   Step 1: Let the width be w meters (the sides perpendicular to the house) and the length be L meters (the side parallel to the house). Since only three sides are fenced, the total fencing is L + 2w = 48. Step 2: Solve for L in terms of w: L = 48 - 2w. Step 3: The area A = length × width = L × w = (48 - 2w) × w = 48w - 2w². Step 4: This is a quadratic function A(w) = -2w² + 48w, which opens downward (coefficient of w² is negative), so it has a maximum at its vertex. Step 5: The vertex occurs at w = -b/(2a) where a = -2 and b = 48. So w = -48/(2 × -2) = -48/(-4) = 12 meters. Step 6: Find L: L = 48 - 2(12) = 48 - 24 = 24 meters. Step 7: Find the maximum area: A = 24 × 12 = 288 square meters. The dimensions that maximize the area are length = 24 m and width = 12 m, giving a maximum area of 288 m².