Quadratic Modeling and Word Problems

6.7: Quadratic Modeling and Word Problems. Original editorial SAT preparation with three worked examples, ten flashcards, and thirty practice questions. Educator-review draft.

Lesson 7 of 83 parts Practice Flashcards

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Area models

Objective: Build a quadratic from two related dimensions.

Prerequisite skills: quadratic functions and contextual modeling. Area of a rectangle is the product of its side lengths. If the sides depend linearly on the same unknown, the area is quadratic. Solve for the side-length variable and reject negative lengths. Common error: adding the dimensions when modeling area. Restrictions and interpretation: Reject roots outside the physical domain; state units for lengths, times, areas, and output values.

Formulas

$A = width × length$

Worked examples

Try itA rectangle is 4 meters longer than it is wide and has area 672 square meters. What is its width?Show answer

Let w be width. Then w(w+4)=672, so w²+4w−672=0. Factor (w−24)(w+28)=0. The roots are 24 and -28; only the positive width 24 meters is feasible. Therefore the requested answer is 24.

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Area models

Objective: Build a quadratic from two related dimensions.

Formulas

$A = width × length$

Worked examples

Try itA rectangle is 4 meters longer than it is wide and has area 672 square meters. What is its width?Show answer

Let w be width. Then w(w+4)=672, so w²+4w−672=0. Factor (w−24)(w+28)=0. The roots are 24 and -28; only the positive width 24 meters is feasible. Therefore the requested answer is 24.

Height over time

Objective: Interpret a quadratic vertex in a motion model.

A downward-opening height model reaches its greatest height at the vertex. Solve for the vertex time, then evaluate height there. Time restrictions and the reference height determine the physically meaningful interval. Common error: returning peak time when the question asks for peak height. Restrictions and interpretation: Reject roots outside the physical domain; state units for lengths, times, areas, and output values.

Formulas

$H(t) = −at² + bt + c ⇒ t_peak = b/(2a), a > 0$

Worked examples

Try itA model gives H(t) = −2t² + 96t + 5 meters for t ≥ 0 until landing. What is the maximum height?Show answer

The peak time is −b/(2a) = −96/(−4) = 24 seconds. Evaluate: H(24) = −2(576) + 2304 + 5 = 1157 meters. This positive time is before landing. Therefore the requested answer is 1157.

Revenue modeling

Objective: Optimize a product of price and demand.

Revenue equals price times quantity, not their sum. A linear demand model often gives a downward-opening quadratic revenue function. Check any feasible price or integer constraints after finding the vertex. Common error: maximizing the demand alone rather than the revenue. Restrictions and interpretation: Reject roots outside the physical domain; state units for lengths, times, areas, and output values. Desmos and method choice: Use Desmos for arithmetic or a graph/table check when that reduces work. Enter parentheses explicitly and verify the requested quantity and units. Use the algebraic steps shown here for an exact result; a rounded graph display is not proof of exact equality.

Formulas

$R(p) = p(A − Bp)$

Worked examples

Try itDemand is q = 112 − 2p units when price p is in dollars, with 0 ≤ p ≤ 56. What price maximizes revenue p·q?Show answer

Revenue is R(p)=p(112−2p)=−2p²+112p. Its vertex is at p=−112/(−4)=28. This lies in the feasible interval; the maximizing price is $28. Therefore the requested answer is 28.

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