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All courses Algebra II · A-CED.3 17 of 55
Represent constraints and systems, then interpret viable and non-viable solutions in modeling contexts.

Write a polynomial constraint for a design or revenue context

Problem
A design has area \(x(20~-~x)\) and requires area at least \(75\). Write the polynomial constraint and feasible design domain.
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Hint

Keep the area expression exactly as given, then translate the phrase \(at least 75\) into an inequality.

A design context usually also needs feasible bounds so the factors in the expression stay positive.

Solution walkthrough

01

Use the supplied area expression

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

The design's two multiplicative dimensions are represented by x and 20-x, so its area is already given as x(20-x).

02

Translate the minimum requirement

\[\text{area}~\text{at}~\text{least}~75~->~A\ge~75\]

The phrase at least includes 75 and every larger value, so it translates to a greater-than-or-equal inequality.

03

Write the polynomial constraint

\[x(20-x)\ge~75\]

Substituting the area expression into the requirement produces the requested polynomial constraint.

04

Derive the feasible design domain

\[x(20-x)~\ge~75~\text{with}~0~<~x~<~20\]

Both design dimensions must be positive, so x must lie strictly between 0 and 20. Therefore the model is x(20-x)>=75 with feasible design domain 0<x<20, choice A.

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Another way

  1. Expand the constraint to -x²+20x>=75 while retaining the separately derived physical domain 0<x<20.

!

Common mistake

Do not use x<=20 and include an endpoint. At x=0 or x=20, one design dimension is zero, so the design is degenerate rather than feasible.

Solution walkthrough video