Quadratic Word Problems

TipLearning Objectives

By the end of this lesson, you’ll be able to:

  • Translate real-world scenarios into quadratic functions.
  • Interpret the meaning of a quadratic’s vertex and roots in context.
  • Solve projectile, area, and revenue/optimization problems using quadratics.

Key Ideas

Many real-world situations naturally lead to quadratic models:

  • Projectile motion: height as a function of time
  • Geometry: area given a fixed perimeter or constraints
  • Economics: revenue, cost, and profit functions

Key interpretations:

  • The vertex gives a maximum or minimum depending on whether the parabola opens down or up.
  • The roots (zeros) often represent start/end times, ground hits, or break-even points.
  • The initial value (plugging in \(0\)) usually gives the starting height, price, or quantity.

Common Problem Types

1. Projectile motion

Use zeros for start/end times and the vertex for maximum height.

2. Area problems

Model with \(A(x) = x(\text{other dimension})\) and find the vertex for max area.

3. Revenue / profit

Use \(R = pq\) or \(P = R - C\), expand into standard form, and maximize using the vertex.

4. Context interpretation

Units matter—always check if the problem wants time, height, price, or quantity.

Strategies

  • Identify what your variable represents before writing your function.

  • Put quadratics into standard form to use:

    \[ \begin{split} x_v &= -\frac{b}{2a} \end{split} \]

  • For projectile problems, set the height to zero to find landing times.

  • For revenue models \(R = (\text{price})(\text{quantity})\), express both in terms of the same variable first.

  • Interpret all results in context: negative time, negative price, or impossible dimensions should be rejected.

Worked Examples

Example 1 — Projectile Motion

A ball is thrown upward from ground level. Its height (meters) after \(t\) seconds is:

\[ h(t) = -5t^2 + 20t \]

(a) When does it hit the ground?
(b) What is the maximum height?

Solution:

(a) Set height to zero:

\[ \begin{split} -5t^2 + 20t &= 0 \\ -5t(t - 4) &= 0 \end{split} \]

So \(t = 0\) (start) or \(t = 4\).
The ball hits the ground at 4 seconds.

(b) Maximum at the vertex:

\[ \begin{split} t_v &= -\frac{b}{2a} \\ &= -\frac{20}{2(-5)} \\ &= 2 \end{split} \]

Now compute height:

\[ \begin{split} h(2) &= -5(4) + 20(2) \\ &= -20 + 40 = 20 \end{split} \]

Maximum height: 20 meters at 2 seconds.

Example 2 — Revenue Optimization

A shop sells an item for $10 and sells 100 units daily. For each $1 increase in price, they sell 10 fewer units. Find the price that maximizes revenue.

Let \(x\) = number of $1 increases.

  • New price: \(10 + x\)
  • New quantity: \(100 - 10x\)

Revenue:

\[ \begin{split} R(x) &= (10 + x)(100 - 10x) \\ &= 1000 - 100x + 100x - 10x^2 \\ &= -10x^2 + 1000 \end{split} \]

Vertex:

\[ x_v = -\frac{b}{2a} = -\frac{0}{2(-10)} = 0 \]

So maximum revenue occurs at no price increase.
Best price: $10.

WarningCommon Mistakes
  • Forgetting what the variable represents.
  • Reporting the vertex’s \(x\)–value when the question asks for a maximum height, profit, or revenue (the \(y\)–value).
  • Giving negative time or negative dimensions.
  • Not interpreting answers in context.

Practice Problems

  1. A projectile is launched from a platform. Height (feet):

    \[ h(t) = -16t^2 + 32t + 48 \]

    1. Initial height
    2. Time of max height
    3. Maximum height

  2. Rectangle area:

    \[ A(x) = x(20 - x) \]

    1. Write in standard form
    2. Value of \(x\) that maximizes area

  3. Dropped ball from 80 m:

    \[ h(t) = -5t^2 + 80 \]

    1. When does it hit the ground?
    2. Time when height = 40 m

  4. Profit model:

    \[ P(q) = -2q^2 + 40q - 96 \]

    1. Quantity maximizing profit
    2. Maximum profit

  5. Basketball shot:

    \[ h(t) = -4t^2 + 12t + 6 \]

    1. Height at \(t = 0\)
    2. Maximum height

1. Projectile:

  1. \[h(0) = 48\]
    Initial height = 48 ft

  2. \[t_v = -\frac{32}{2(-16)} = 1\]
    Max height occurs at 1 sec

  3. \[h(1) = -16 + 32 + 48 = 64\]
    Maximum height = 64 ft

2. Rectangle:

  1. \[A(x) = x(20 - x) = -x^2 + 20x\]

  2. \[x_v = -\frac{20}{2(-1)} = 10\]
    Area maximized at 10 m

3. Dropped ball:

  1. \[-5t^2 + 80 = 0\]
    \[t^2 = 16 \Rightarrow t = 4\]
    Hits ground at 4 sec

  2. \[-5t^2 + 80 = 40\]
    \[-5t^2 + 40 = 0\]
    \[t^2 = 8 \Rightarrow t = \sqrt{8} = 2\sqrt{2}\]
    Time ≈ 2.83 sec

4. Profit:

  1. \[q_v = -\frac{40}{2(-2)} = 10\]
    Profit maximized at 10 items

  2. \[P(10) = -200 + 400 - 96 = 104\]
    Max profit = $104

5. Basketball shot:

  1. \[h(0) = 6\]
    Initial height = 6

  2. \[t_v = -\frac{12}{2(-4)} = \frac{3}{2}\]
    \[h\!\left(\frac{3}{2}\right) = -9 + 18 + 6 = 15\]
    Maximum height = 15

Summary

  • Use quadratics to model projectile height, area, and revenue/profit.
  • The vertex gives maximum or minimum values depending on direction of opening.
  • Roots represent times or quantities where the value hits zero.
  • Always interpret results logically: time, height, length, and revenue must make sense.
  • Identify the variable and what it means before writing equations.
  • Use \(x_v = -\frac{b}{2a}\) for quick optimization.
  • Context matters: ignore negative or impossible values.
  • Vertex → max/min; roots → start/end/break-even.