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Quadratic Applications

A free Algebra I lesson from the “Quadratic Foundations” unit, with a worked example and practice problems including step-by-step solutions.

A quadratic model often arises when two changing quantities are multiplied, as with the sides of a rectangle, or when a height is modeled by a squared time term. Define the variable and its allowed values before solving. A zero answers when the modeled quantity equals zero; a vertex answers where it is largest or smallest. These are different questions. Translate the requested condition into an equation, solve it, and then return to the context: a negative length is not a physical solution, and time after an object's impact is outside a simple flight model. For a fixed-perimeter rectangle, writing one side in terms of the other reduces the area to one variable. Completing the square then proves a maximum, since the square is nonnegative. For a height model, height units come from the whole expression and time units from the input. A calculated optimum is justified only under the model's assumptions, such as fixed perimeter, constant prices, or an idealized flight. Check the original constraints as well as the algebra.

What you'll learn

Why it matters: A garden planner can maximize area with a limited fence, while a flight model uses the vertex for peak height and a zero for landing time.

Worked example

Problem. A rectangular garden uses 24 meters of fencing on all four sides. Let w be its width. Model its area and find the dimensions that maximize it.

  1. Let l be the length. The perimeter condition is 2w + 2l = 24, so l = 12 - w.
  2. Both dimensions must be positive, requiring 0 < w < 12.
  3. Area is width times length: A(w) = w(12 - w) = 12w - w^2 square meters.
  4. Complete the square: A(w) = 36 - (w - 6)^2.
  5. A nonnegative square is subtracted from 36, so area cannot exceed 36; equality occurs at w = 6, which is permitted.
  6. Then l = 12 - 6 = 6. Check perimeter 2(6) + 2(6) = 24 and area 6(6) = 36.

Answer: A(w) = 12w - w^2 for 0 < w < 12; maximum 36 square meters at width 6 meters and length 6 meters.

Lesson preview: this page shows 5 practice items. The interactive activity contains 32 practice items and 17 quiz items (typed response and multiple choice). Counts include repeated prompts and variants.

Practice problems

1. A rectangular garden uses 40 meters of fencing on all four sides. What is its greatest possible area in square meters?

Show solution
  1. If width is w, length is 20 - w, so A(w) = 20w - w^2.
  2. A(w) = 100 - (w - 10)^2, which is at most 100.
  3. Width and length both 10 meters attain that area with perimeter 40.

Answer: 100

2. A ball's height in meters is h(t) = -5t^2 + 20t + 1 for its flight after t = 0. What is its maximum height?

Show solution
  1. Complete the square: h(t) = -5(t - 2)^2 + 21.
  2. The squared term is smallest at t = 2, so height is largest there.
  3. h(2) = -20 + 40 + 1 = 21 meters.

Answer: 21

3. A rectangle has width w and length w + 3 with area 40. Solving gives w = 5 or w = -8. What are the physical dimensions?

Choices: Width 5 and length 8 · Width -8 and length -5 · Width 5 and length -8 · Both pairs represent physical rectangles.

Show solution
  1. The model w(w + 3) = 40 gives (w + 8)(w - 5) = 0.
  2. Negative dimensions are outside the physical domain, so w = -8 is rejected.
  3. For w = 5, length is 8 and area is 40.

Answer: Width 5 and length 8

4. A ball follows h(t) = -t^2 + 6t from launch at t = 0 until it returns to the ground. Which graph feature answers 'When is the ball highest?'

Choices: The vertex input t = 3 · The later zero t = 6 · The height output h = 0 · The initial input t = 0

Show solution
  1. Rewrite h(t) = 9 - (t - 3)^2.
  2. The vertex is (3, 9), meaning a maximum height of 9 at time 3.
  3. The later zero at t = 6 answers the different question of landing time.

Answer: The vertex input t = 3

5. A rectangle has area x(x + 5). If x = 4, what is the area?

Show solution
  1. Substitute x = 4.
  2. 4(9) = 36.

Answer: 36

Explain and apply

Complete these tasks on paper. Use the rubric to review your reasoning with a teacher or study partner; these tasks are not automatically scored.

Model an enclosure with a missing fence side

A straight wall forms one side of a rectangular enclosure. You have 30 meters of fencing for the other three sides. Define a variable for the perpendicular sides, build an area model, and prove which dimensions give the maximum area. Explain why the four-sided-perimeter formula would be wrong here.

Review rubric and solution
  • Let w be each perpendicular side and l the side parallel to the wall, with 2w + l = 30.
  • Obtain A(w) = w(30 - 2w), with 0 < w < 15.
  • Complete the square to A(w) = 112.5 - 2(w - 7.5)^2.
  • Conclude dimensions 7.5 by 15 meters and maximum area 112.5 square meters; explain that the wall side uses no fencing.
  1. Only three sides use fencing, so l = 30 - 2w rather than 15 - w.
  2. Positive dimensions require 0 < w < 15. Area is A(w) = 30w - 2w^2.
  3. Rewrite A(w) = -2(w^2 - 15w) = 112.5 - 2(w - 7.5)^2.
  4. The maximum is 112.5 square meters when w = 7.5 and l = 15. The fence total is 7.5 + 7.5 + 15 = 30 meters.
  5. Using 2w + 2l = 30 would incorrectly allocate fencing to the wall side.

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