Applications of Polynomial Functions
Overview
Polynomial functions have a wide range of applications in various fields, including physics, engineering, economics, and biology. They can model real-world situations where relationships between variables follow a smooth, continuous pattern. Common applications of polynomials include:
- Projectile Motion: The height of an object in motion under gravity can be modeled by a quadratic function, such as \( h(t) = -16t^2 + v_0 t + h_0 \), where \( v_0 \) is the initial velocity and \( h_0 \) is the initial height.
- Optimization Problems: Polynomial functions are often used to model cost, revenue, and profit functions. These functions can be optimized to maximize or minimize certain quantities.
- Population Growth: In some cases, population growth can be modeled using polynomial functions, where the degree of the polynomial indicates the rate of change over time.
- Economics and Finance: Polynomial functions are used to model supply and demand curves, pricing strategies, and even stock market trends.
Example 1: Projectile Motion
Consider the height of a ball thrown into the air modeled by the quadratic function:
\( h(t) = -16t^2 + 40t + 5 \)
Where \( h(t) \) is the height in feet and \( t \) is the time in seconds. The equation models the height of the ball at any given time. To find the time at which the ball reaches its maximum height, we can find the vertex of the parabola:
\( t = -\frac{b}{2a} = -\frac{40}{2(-16)} = 1.25 \) seconds
The ball reaches its maximum height of:
\( h(1.25) = -16(1.25)^2 + 40(1.25) + 5 = 35 \) feet
Example 2: Optimizing a Revenue Function
Suppose the revenue \( R(x) \) from selling \( x \) items is given by the polynomial function:
\( R(x) = -2x^2 + 40x \)
To find the number of items \( x \) that maximizes the revenue, we can find the vertex of the quadratic function:
\( x = -\frac{b}{2a} = -\frac{40}{2(-2)} = 10 \) items
The maximum revenue is:
\( R(10) = -2(10)^2 + 40(10) = 200 \) dollars
Practice Questions
- Question 1: A rocket is launched with an initial velocity of 48 m/s and an initial height of 100 meters. The height \( h(t) \) of the rocket at time \( t \) is modeled by the function:
\( h(t) = -4.9t^2 + 48t + 100 \)
Find the time at which the rocket reaches its maximum height and the maximum height.
Solution
To find the time at which the rocket reaches its maximum height, we use the vertex formula:
\( t = -\frac{b}{2a} = -\frac{48}{2(-4.9)} \approx 4.9 \) seconds
The maximum height is:
\( h(4.9) = -4.9(4.9)^2 + 48(4.9) + 100 \approx 324.4 \) meters
- Question 2: The profit \( P(x) \) from selling \( x \) units of a product is modeled by the function:
\( P(x) = -3x^2 + 60x - 200 \)
Find the number of units \( x \) that maximizes the profit and the maximum profit.
Solution
To find the number of units that maximizes the profit, we use the vertex formula:
\( x = -\frac{b}{2a} = -\frac{60}{2(-3)} = 10 \) units
The maximum profit is:
\( P(10) = -3(10)^2 + 60(10) - 200 = 400 \) dollars
- Question 3: The height \( h(t) \) of a projectile is given by the equation:
\( h(t) = -16t^2 + 64t + 80 \)
Find the time at which the projectile reaches its maximum height and the maximum height.
Solution
Using the vertex formula to find the time:
\( t = -\frac{b}{2a} = -\frac{64}{2(-16)} = 2 \) seconds
The maximum height is:
\( h(2) = -16(2)^2 + 64(2) + 80 = 192 \) feet
- Question 4: A company manufactures and sells \( x \) units of a product, and its total cost function is given by:
\( C(x) = 0.5x^2 - 10x + 200 \)
Find the number of units \( x \) that minimizes the cost and the minimum cost.
Solution
Using the vertex formula to find the number of units that minimizes the cost:
\( x = -\frac{b}{2a} = -\frac{-10}{2(0.5)} = 10 \) units
The minimum cost is:
\( C(10) = 0.5(10)^2 - 10(10) + 200 = 150 \) dollars
- Question 5: The revenue function for a company is modeled by the polynomial:
\( R(x) = -x^2 + 12x \)
Find the number of items \( x \) that maximizes the revenue and the maximum revenue.
Solution
To find the number of items that maximizes the revenue, we use the vertex formula:
\( x = -\frac{b}{2a} = -\frac{12}{2(-1)} = 6 \) items
The maximum revenue is:
\( R(6) = -(6)^2 + 12(6) = 36 \) dollars