Quadratic Modeling

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Quadratic Modeling · Grade 10 · Worksheet 2

1. Matiu is a farmer who grows organic strawberries. He notices that the number of strawberry plants he can sell each week, n, depends on the price per plant, p dollars, according to the relationship n = 2400 - 60p. The cost to grow and package each plant is $8, and he has fixed weekly costs of $3200. Determine the price Matiu should charge per plant to maximize his weekly profit, and what is that maximum weekly profit? Answer: Price = $24, Maximum profit = $6400 Solution: Express the number of plants sold n in terms of price p: n = 2400 - 60p. Revenue R = price per plant times number sold = p * n = p(2400 - 60p) = 2400p - 60p^2.
   Full step-by-step solution

   Step 1: Express the number of plants sold n in terms of price p: n = 2400 - 60p. Step 2: Revenue R = price per plant times number sold = p * n = p(2400 - 60p) = 2400p - 60p^2. Step 3: Variable cost = cost per plant times number sold = 8n = 8(2400 - 60p) = 19200 - 480p. Step 4: Total cost C = variable cost + fixed cost = (19200 - 480p) + 3200 = 22400 - 480p. Step 5: Profit P = Revenue - Cost = (2400p - 60p^2) - (22400 - 480p) = 2400p - 60p^2 - 22400 + 480p = -60p^2 + 2880p - 22400. Step 6: This is a quadratic function in the form P(p) = -60p^2 + 2880p - 22400. It opens downward, so the vertex gives the maximum profit. Step 7: The vertex occurs at p = -b/(2a) where a = -60 and b = 2880. So p = -2880/(2*(-60)) = -2880/(-120) = 24. Step 8: The price that maximizes profit is $24 per plant. Step 9: Substitute p = 24 into the profit function: P(24) = -60(24)^2 + 2880(24) - 22400 = -60(576) + 69120 - 22400 = -34560 + 69120 - 22400 = 12160. Step 10: The maximum weekly profit is $12,160. The answer is: Price = $24, Maximum profit = $12,160.

2. A projectile is launched from ground level. Its height h (in meters) at time t (in seconds) is modeled by the quadratic function h(t) = -5t² + 30t. What is the maximum height reached by the projectile? Answer: 45 Solution: We are given the height function: h(t) = -5t² + 30t. Recognize the type of function. This is a quadratic function in the form h(t) = at² + bt + c, where a = -5, b = 30, c = 0.
   Full step-by-step solution

   We are given the height function: h(t) = -5t² + 30t. Step 1: Recognize the type of function. This is a quadratic function in the form h(t) = at² + bt + c, where a = -5, b = 30, c = 0. Since a is negative, the parabola opens downward, so the vertex gives the maximum height. Step 2: Find the time at which maximum height occurs. For a quadratic function at² + bt + c, the t-coordinate of the vertex is given by: t = -b / (2a) Substitute b = 30 and a = -5: t = -30 / (2 * -5) = -30 / (-10) = 3. So the projectile reaches its maximum height at t = 3 seconds. Step 3: Find the maximum height by evaluating h(3). h(3) = -5*(3)² + 30*(3) = -5*9 + 90 = -45 + 90 = 45. Step 4: Conclusion. The maximum height reached by the projectile is 45 meters.

3. Isabella is designing a rectangular garden next to a straight river. She has 72 meters of fencing to enclose the other three sides of the garden (the side along the river needs no fencing). She wants to maximize the area of the garden. What dimensions (length parallel to the river and width perpendicular to the river) will give the maximum area, and what is that maximum area? Answer: Dimensions: 36 m by 18 m; Maximum area: 648 m² Solution: Let the width perpendicular to the river be w meters, and the length parallel to the river be L meters. Only three sides need fencing: one length and two widths. So L + 2w = 72, which gives L = 72 - 2w.
   Full step-by-step solution

   Let the width perpendicular to the river be w meters, and the length parallel to the river be L meters. Only three sides need fencing: one length and two widths. So L + 2w = 72, which gives L = 72 - 2w. The area A = L * w = (72 - 2w) * w = 72w - 2w². This is a quadratic in standard form: A = -2w² + 72w. The vertex (maximum) occurs at w = -b/(2a) where a = -2, b = 72. So w = -72/(2 * -2) = -72/-4 = 18. Then L = 72 - 2*18 = 72 - 36 = 36. The maximum area is A = 36 * 18 = 648 m². So the dimensions are 36 m (parallel to river) by 18 m (perpendicular), and the maximum area is 648 square meters.

4. A parabolic arch bridge has its vertex at (0, 20) and spans a horizontal distance of 40 meters, with the endpoints at (-20, 0) and (20, 0). The arch follows the path of a quadratic function. What is the height of the arch 10 meters horizontally from the center? Answer: 15 Solution: - Vertex at (0, 20) - Ends at (-20, 0) and (20, 0) - Quadratic function: y = a(x - h)^2 + k Vertex (h, k) = (0, 20) y = a(x - 0)^2 + 20 y = a x^2 + 20 Use point (20, 0): 0 = a(20)^2 + 20 0 = 400a + 20 400a = -20 a = -20/400 a = -1/20 y = (-1/20) x^2 + 20 Find height 10 meters from center "10…
   Full step-by-step solution

   Let's solve step-by-step. --- **Step 1: Understand the problem** We have a parabolic arch with: - Vertex at (0, 20) - Ends at (-20, 0) and (20, 0) - Quadratic function: y = a(x - h)^2 + k --- **Step 2: Write the vertex form** Vertex (h, k) = (0, 20) So: y = a(x - 0)^2 + 20 y = a x^2 + 20 --- **Step 3: Use a point to find 'a'** Use point (20, 0): 0 = a(20)^2 + 20 0 = 400a + 20 400a = -20 a = -20/400 a = -1/20 --- **Step 4: Write the equation** y = (-1/20) x^2 + 20 --- **Step 5: Find height 10 meters from center** "10 meters horizontally from center" means x = 10 or x = -10. Let's take x = 10: y = (-1/20)(10)^2 + 20 y = (-1/20)(100) + 20 y = -100/20 + 20 y = -5 + 20 y = 15 --- **Step 6: Conclusion** The height of the arch 10 meters from the center is 15 meters. **Final answer: 15**

5. Hana launches a model rocket from the ground. The height h(t) in meters after t seconds is modeled by a quadratic function. The rocket reaches a maximum height of 144 meters at 6 seconds. Find the height of the rocket after 4 seconds. Answer: 128 Solution: The vertex form is h(t) = a(t - 6)^2 + 144. At t = 0, h(0) = 0, so 0 = a(0 - 6)^2 + 144. Simplify: 0 = a(36) + 144, so 36a = -144, thus a = -4.
   Full step-by-step solution

   Step 1: The vertex form is h(t) = a(t - 6)^2 + 144. Step 2: At t = 0, h(0) = 0, so 0 = a(0 - 6)^2 + 144. Step 3: Simplify: 0 = a(36) + 144, so 36a = -144, thus a = -4. Step 4: The model is h(t) = -4(t - 6)^2 + 144. Step 5: At t = 4: h(4) = -4(4 - 6)^2 + 144 = -4(-2)^2 + 144 = -4(4) + 144 = -16 + 144 = 128. The height after 4 seconds is 128 meters.

6. 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.

7. A rectangular garden has an area of 60 m². If the length is 4 meters more than the width, find the dimensions of the garden. (Let w represent the width) Answer: w = 6, l = 10 Solution: Let w = width, then length = w + 4 Area = length × width = (w + 4) × w = 60 Expand: w² + 4w = 60 Rearrange: w² + 4w - 60 = 0 Factor: (w + 10)(w - 6) = 0 Solve: w = -10 or w = 6 Since width cannot be negative, w = 6 Length = w + 4 = 6 + 4 = 10 Check: 6 × 10 = 60 m² The dimensions are width = 6 m…
   Full step-by-step solution

   Step 1: Let w = width, then length = w + 4 Step 2: Area = length × width = (w + 4) × w = 60 Step 3: Expand: w² + 4w = 60 Step 4: Rearrange: w² + 4w - 60 = 0 Step 5: Factor: (w + 10)(w - 6) = 0 Step 6: Solve: w = -10 or w = 6 Step 7: Since width cannot be negative, w = 6 Step 8: Length = w + 4 = 6 + 4 = 10 Step 9: Check: 6 × 10 = 60 m² The dimensions are width = 6 m and length = 10 m.