2.6 Modelling With Quadratics
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Learning Objectives
- Understand how quadratic functions can be used to model real-world situations
- Formulate quadratic models from given information
- Interpret the key features of quadratic models in context
- Solve optimization problems using quadratic models
- Recognize the limitations of quadratic models
Key Concepts
Introduction to Mathematical Modelling
Mathematical modelling is the process of using mathematics to represent and analyze real-world situations.
Quadratic functions are particularly useful for modelling situations where a quantity increases (or decreases) to a maximum (or minimum) value and then decreases (or increases).
Examples include projectile motion, profit functions, area problems, and many other applications in physics, economics, engineering, and other fields.
Formulating Quadratic Models
To formulate a quadratic model, we need to identify the variables and the relationship between them.
Often, we need to translate verbal descriptions or data into a mathematical equation.
The general form of a quadratic function is f(x) = ax² + bx + c, where a, b, and c are constants and a ≠ 0.
Example 1
A rectangular garden has a perimeter of 40 meters. Express the area A of the garden as a function of the width x.
Solution:
Step 1: Define the variables. Let x be the width of the garden in meters, and let y be the length in meters. Step 2: Use the perimeter constraint. The perimeter is 40 meters, so 2x + 2y = 40. Solving for y, we get y = 20 - x. Step 3: Express the area as a function of x. The area is A = xy = x(20 - x) = 20x - x². So the area function is A(x) = -x² + 20x, which is a quadratic function with a = -1, b = 20, and c = 0.
Example 2
A company finds that the cost C (in dollars) of producing x units of a product is given by C = 2000 + 10x + 0.01x². The revenue R (in dollars) from selling x units is R = 20x. Express the profit P as a function of x.
Solution:
Step 1: Define the profit function. Profit is revenue minus cost: P = R - C. Step 2: Substitute the given expressions for R and C. P = 20x - (2000 + 10x + 0.01x²) P = 20x - 2000 - 10x - 0.01x² P = -0.01x² + 10x - 2000 So the profit function is P(x) = -0.01x² + 10x - 2000, which is a quadratic function with a = -0.01, b = 10, and c = -2000.
Interpreting Key Features in Context
The key features of a quadratic function (vertex, intercepts, etc.) often have meaningful interpretations in the context of the problem.
The vertex represents the maximum or minimum value of the function, which can correspond to optimal solutions in many applications.
The x-intercepts represent the values of the independent variable where the dependent variable equals zero, which can have specific meanings in context.
Example 1
For the garden problem above, where A(x) = -x² + 20x, interpret the key features of the quadratic function in context.
Solution:
Step 1: Find the vertex of the quadratic function. Since a = -1 < 0, the parabola opens downward, and the vertex is a maximum. Using the formula x = -b/(2a), the x-coordinate of the vertex is x = -20/(2(-1)) = 10. The maximum area is A(10) = -(10)² + 20(10) = -100 + 200 = 100 square meters. Step 2: Find the x-intercepts. Setting A(x) = 0, we get -x² + 20x = 0, which gives x(20 - x) = 0, so x = 0 or x = 20. Step 3: Interpret these features in context. - The maximum area of 100 square meters is achieved when the width is 10 meters (and consequently, the length is also 10 meters, making it a square). - When x = 0, the area is 0, which means the garden has zero width (and is just a line). - When x = 20, the area is also 0, which means the garden has zero length (and is just a line). - For 0 < x < 20, the area is positive, which represents physically meaningful garden dimensions.
Example 2
For the profit function P(x) = -0.01x² + 10x - 2000, interpret the key features in context.
Solution:
Step 1: Find the vertex of the quadratic function. Since a = -0.01 < 0, the parabola opens downward, and the vertex is a maximum. Using the formula x = -b/(2a), the x-coordinate of the vertex is x = -10/(2(-0.01)) = 500. The maximum profit is P(500) = -0.01(500)² + 10(500) - 2000 = -0.01(250000) + 5000 - 2000 = -2500 + 5000 - 2000 = 500 dollars. Step 2: Find the x-intercepts (break-even points). Setting P(x) = 0, we get -0.01x² + 10x - 2000 = 0. Using the quadratic formula with a = -0.01, b = 10, c = -2000: x = (-10 ± √(100 - 4(-0.01)(-2000)))/(-0.02) = (-10 ± √(100 - 80))/(-0.02) = (-10 ± √20)/(-0.02) x ≈ (-10 + 4.47)/(-0.02) ≈ 277.3 or x ≈ (-10 - 4.47)/(-0.02) ≈ 722.7 Step 3: Interpret these features in context. - The maximum profit of $500 is achieved when producing and selling 500 units. - The company breaks even (P = 0) when producing and selling approximately 277 units or 723 units. - For production levels between 277 and 723 units, the company makes a profit (P > 0). - For production levels below 277 units or above 723 units, the company operates at a loss (P < 0).
Optimization Problems
Optimization problems involve finding the maximum or minimum value of a function, often subject to constraints.
For quadratic functions, the maximum or minimum occurs at the vertex of the parabola.
Many real-world problems involve finding optimal values, such as maximizing profit, minimizing cost, or finding the most efficient dimensions.
Example 1
A farmer has 1000 meters of fencing to enclose a rectangular field. What dimensions will maximize the area of the field?
Solution:
Step 1: Define the variables. Let x be the width of the field in meters, and let y be the length in meters. Step 2: Use the perimeter constraint. The perimeter is 1000 meters, so 2x + 2y = 1000. Solving for y, we get y = 500 - x. Step 3: Express the area as a function of x. The area is A = xy = x(500 - x) = 500x - x². Step 4: Find the maximum area by finding the vertex of the quadratic function A(x) = -x² + 500x. Since a = -1 < 0, the parabola opens downward, and the vertex is a maximum. Using the formula x = -b/(2a), the x-coordinate of the vertex is x = -500/(2(-1)) = 250. The maximum area is A(250) = -(250)² + 500(250) = -62500 + 125000 = 62500 square meters. Step 5: Find the corresponding value of y. When x = 250, y = 500 - 250 = 250. So the dimensions that maximize the area are 250 meters by 250 meters, giving a square field with an area of 62500 square meters.
Example 2
A manufacturer finds that the daily cost C (in dollars) of producing x units of a product is given by C = 0.02x² + 10x + 500. What production level minimizes the average cost per unit?
Solution:
Step 1: Define the average cost function. The average cost per unit is C/x. Step 2: Express the average cost as a function of x. C/x = (0.02x² + 10x + 500)/x = 0.02x + 10 + 500/x Step 3: Find the minimum average cost. This is not a quadratic function, but we can find the minimum by taking the derivative and setting it equal to zero. Let A(x) = 0.02x + 10 + 500/x be the average cost function. A'(x) = 0.02 - 500/x² Setting A'(x) = 0, we get 0.02 = 500/x², which gives x² = 500/0.02 = 25000, so x = 158.11. Step 4: Verify that this is a minimum by checking that A''(x) > 0. A''(x) = 1000/x³, which is positive for x > 0. So the production level that minimizes the average cost per unit is approximately 158 units per day.
Projectile Motion
One of the most common applications of quadratic functions is in modelling the motion of projectiles under the influence of gravity.
The height h of a projectile at time t is given by h = h0 + v0*t - (1/2)g*t^2, where h0 is the initial height, v0 is the initial vertical velocity, and g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth).
This is a quadratic function of time, and its key features have physical interpretations.
Example 1
A ball is thrown vertically upward from a height of 1.5 meters with an initial velocity of 20 m/s. Find (a) the maximum height reached by the ball, (b) the time it takes to reach the maximum height, and (c) the time it takes for the ball to hit the ground.
Solution:
Step 1: Set up the height function. Using the formula h = h0 + v0*t - (1/2)g*t^2, with h0 = 1.5 m, v0 = 20 m/s, and g = 9.8 m/s^2, we get h = 1.5 + 20t - 4.9t^2. Step 2: Find the time to reach the maximum height. The maximum height occurs when the velocity is zero, which happens when dh/dt = 0. We have dh/dt = 20 - 9.8t, so 20 - 9.8t = 0 gives t = 20/9.8 ≈ 2.04 seconds. Step 3: Find the maximum height. The maximum height is h(2.04) = 1.5 + 20(2.04) - 4.9(2.04)^2 ≈ 1.5 + 40.8 - 20.4 ≈ 21.9 meters. Step 4: Find the time when the ball hits the ground. This happens when h = 0, so we solve 1.5 + 20t - 4.9t^2 = 0. Using the quadratic formula with a = -4.9, b = 20, c = 1.5: t = (-20 ± √(400 - 4(-4.9)(1.5)))/(-9.8) = (-20 ± √(400 + 29.4))/(-9.8) = (-20 ± √429.4)/(-9.8) t ≈ (-20 + 20.72)/(-9.8) ≈ -0.07 seconds (invalid) or t ≈ (-20 - 20.72)/(-9.8) ≈ 4.16 seconds So the ball reaches a maximum height of about 21.9 meters after 2.04 seconds, and it hits the ground after about 4.16 seconds.
Example 2
A projectile is fired from ground level with an initial velocity of 50 m/s at an angle of 30° to the horizontal. Find (a) the maximum height reached by the projectile, (b) the time it takes to reach the maximum height, and (c) the horizontal distance traveled before the projectile returns to the ground (the range).
Solution:
Step 1: Resolve the initial velocity into horizontal and vertical components. Horizontal component: v0x = 50 cos(30°) = 50 × 0.866 ≈ 43.3 m/s Vertical component: v0y = 50 sin(30°) = 50 × 0.5 = 25 m/s Step 2: Set up the position functions. The horizontal position is x = v0x*t = 43.3t, and the vertical position (height) is y = v0y*t - (1/2)g*t^2 = 25t - 4.9t^2. Step 3: Find the time to reach the maximum height. The maximum height occurs when the vertical velocity is zero, which happens when dy/dt = 0. We have dy/dt = 25 - 9.8t, so 25 - 9.8t = 0 gives t = 25/9.8 ≈ 2.55 seconds. Step 4: Find the maximum height. The maximum height is y(2.55) = 25(2.55) - 4.9(2.55)^2 ≈ 63.75 - 31.88 ≈ 31.87 meters. Step 5: Find the time when the projectile returns to the ground. This happens when y = 0, so we solve 25t - 4.9t^2 = 0. Factoring out t, we get t(25 - 4.9t) = 0, which gives t = 0 or t = 25/4.9 ≈ 5.1 seconds. Since t = 0 corresponds to the initial position, the projectile returns to the ground after about 5.1 seconds. Step 6: Find the range. The horizontal distance traveled is x = 43.3t = 43.3 × 5.1 ≈ 220.83 meters. So the projectile reaches a maximum height of about 31.87 meters after 2.55 seconds, and it travels a horizontal distance of about 220.83 meters before returning to the ground.
Limitations of Quadratic Models
While quadratic models are useful for many applications, they have limitations and may not accurately represent all real-world situations.
Quadratic models assume that the rate of change of the rate of change (the second derivative) is constant, which may not be true in complex systems.
It's important to validate the model by comparing its predictions with actual data and to be aware of the range of validity of the model.
Example 1
A company models its daily profit P (in dollars) as a function of the number of units x produced and sold using the quadratic function P = -0.01x² + 10x - 2000. Discuss the limitations of this model.
Solution:
Limitations of the quadratic profit model: 1. The model assumes that the price per unit and the cost per unit follow specific patterns that result in a quadratic profit function. In reality, prices might need to be lowered to sell more units, and costs might have economies or diseconomies of scale that don't follow a quadratic pattern. 2. The model predicts that profit will eventually become negative for large enough values of x, which might not be realistic if the company can adjust its operations for large production volumes. 3. The model assumes that fractional units can be produced and sold, which might not be realistic for certain products. 4. The model doesn't account for external factors such as market conditions, competition, or seasonal variations that might affect profit. 5. The model has a limited range of validity. For example, it predicts that producing zero units results in a loss of $2000, which represents fixed costs. But if the company were to shut down completely, it might be able to eliminate some of these costs. To improve the model, the company could collect more data points, consider using piecewise functions for different production ranges, or incorporate additional variables such as time, market conditions, or production capacity.
Example 2
The height h (in meters) of a projectile t seconds after being launched is modelled by the function h = -4.9t² + 20t + 1.5. Discuss the limitations of this model.
Solution:
Limitations of the quadratic projectile motion model: 1. The model assumes that the only force acting on the projectile is gravity, ignoring air resistance. In reality, air resistance can significantly affect the trajectory, especially for lightweight objects or at high speeds. 2. The model assumes that the acceleration due to gravity is constant at 9.8 m/s^2. In reality, the gravitational acceleration varies slightly with altitude and geographical location. 3. The model assumes that the motion occurs in a vertical plane. If there are crosswinds or if the projectile is spinning, the actual trajectory might deviate from the predicted path. 4. The model assumes that the initial conditions (height and velocity) are known exactly. In practice, there might be measurement errors or variations in the launch conditions. 5. The model doesn't account for other factors that might affect the motion, such as the shape and size of the projectile, weather conditions, or the rotation of the Earth (Coriolis effect). For more accurate predictions, especially for long-range trajectories or in situations where air resistance is significant, more complex models that incorporate these additional factors would be needed.
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