Quadratic Word Problems
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.
- 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
A projectile is launched from a platform. Height (feet):
\[ h(t) = -16t^2 + 32t + 48 \]
- Initial height
- Time of max height
- Maximum height
- Initial height
Rectangle area:
\[ A(x) = x(20 - x) \]
- Write in standard form
- Value of \(x\) that maximizes area
- Write in standard form
Dropped ball from 80 m:
\[ h(t) = -5t^2 + 80 \]
- When does it hit the ground?
- Time when height = 40 m
- When does it hit the ground?
Profit model:
\[ P(q) = -2q^2 + 40q - 96 \]
- Quantity maximizing profit
- Maximum profit
- Quantity maximizing profit
Basketball shot:
\[ h(t) = -4t^2 + 12t + 6 \]
- Height at \(t = 0\)
- Maximum height
- Height at \(t = 0\)
1. Projectile:
\[h(0) = 48\]
Initial height = 48 ft\[t_v = -\frac{32}{2(-16)} = 1\]
Max height occurs at 1 sec\[h(1) = -16 + 32 + 48 = 64\]
Maximum height = 64 ft
2. Rectangle:
\[A(x) = x(20 - x) = -x^2 + 20x\]
\[x_v = -\frac{20}{2(-1)} = 10\]
Area maximized at 10 m
3. Dropped ball:
\[-5t^2 + 80 = 0\]
\[t^2 = 16 \Rightarrow t = 4\]
Hits ground at 4 sec\[-5t^2 + 80 = 40\]
\[-5t^2 + 40 = 0\]
\[t^2 = 8 \Rightarrow t = \sqrt{8} = 2\sqrt{2}\]
Time ≈ 2.83 sec
4. Profit:
\[q_v = -\frac{40}{2(-2)} = 10\]
Profit maximized at 10 items\[P(10) = -200 + 400 - 96 = 104\]
Max profit = $104
5. Basketball shot:
\[h(0) = 6\]
Initial height = 6\[t_v = -\frac{12}{2(-4)} = \frac{3}{2}\]
\[h\!\left(\frac{3}{2}\right) = -9 + 18 + 6 = 15\]
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.