Applications of Quadratics in Real Life

Quadratic equations are not just abstract mathematical concepts—they have many real-world applications! Quadratic functions are used in physics, engineering, finance, architecture, and many other fields to model situations where relationships follow a curved or parabolic pattern.

Common Applications

  • Projectile Motion: When an object is thrown or launched, its path often follows a parabolic trajectory. Quadratic equations help calculate maximum height, time of flight, and range.
  • Area Optimization: Quadratic equations can optimize areas within specific constraints, like maximizing the area enclosed by a given perimeter.
  • Profit Maximization: In business, quadratic functions model revenue and profit scenarios to help maximize earnings or minimize costs.

Example: Projectile Motion

Consider a ball thrown upward from a height of 2 meters with an initial velocity of 20 meters per second. Its height \( h \) (in meters) after \( t \) seconds can be described by the quadratic equation:

$$ h = -5t^2 + 20t + 2 $$

This equation can help us determine how long it takes for the ball to reach the ground, its maximum height, and when it reaches that height.

Practice Questions

  1. A basketball player throws a ball with an initial upward velocity of 15 meters per second from a height of 3 meters. The ball’s height \( h \) (in meters) after \( t \) seconds is given by:

    $$ h = -5t^2 + 15t + 3 $$

    Find:

    • The maximum height the ball reaches.
    • The time it takes for the ball to reach the maximum height.
    • The time it takes for the ball to hit the ground.
    Solution

    Rewrite the equation to find maximum height by completing the square or using vertex form:

    The maximum height is 11.25 meters, reached at \( t = 1.5 \) seconds.

    Solving \( -5t^2 + 15t + 3 = 0 \) for \( t \) gives \( t \approx 3.2 \) seconds when the ball hits the ground.

  2. A company finds that its profit \( P \) (in dollars) from producing \( x \) units of a product is given by the equation:

    $$ P = -2x^2 + 40x - 150 $$

    Determine:

    • The number of units the company should produce to maximize profit.
    • The maximum profit.
    Solution

    The maximum profit occurs at the vertex of the parabola. For a quadratic \( ax^2 + bx + c \), the vertex \( x \)-value is given by \( x = -\frac{b}{2a} \):

    Here, \( x = 10 \) units gives maximum profit.

    Substitute \( x = 10 \) into the equation: \( P = -2(10)^2 + 40(10) - 150 = 50 \).

    Maximum profit is $50.

  3. A farmer wants to fence a rectangular garden along the side of a barn. The barn side is already fenced, so only three sides need fencing. If the farmer has 60 meters of fencing available and wants to maximize the area, what dimensions should the garden have?
    Solution

    Let the width be \( x \). Then, the length is \( (60 - 2x) \), and the area \( A \) is:

    $$ A = x(60 - 2x) = -2x^2 + 60x $$

    Maximize \( A \) by finding the vertex: \( x = 15 \).

    The optimal dimensions are 15 meters by 30 meters.

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