6. Modeling and Context

Nonlinear systems can model situations where one relationship is linear and another has a squared term.

Example: A ball's height is modeled by \(h=-t^2+6t+4\). A platform's height is \(h=12\). To find when the ball is at platform height, solve:
\[ -t^2+6t+4=12. \]
Move everything to one side:
\[ t^2-6t+8=0. \]
Factor:
\[ (t-2)(t-4)=0. \]
The ball is at height 12 at \(t=2\) and \(t=4\). In context, two times can make sense: going up and coming down.

A practical decision tree:

• If both equations equal \(y\), set them equal.
• If one equation gives \(x\) or \(y\), substitute.
• If the question asks "how many solutions," use the discriminant.
• If a square equals a positive number, remember two possible roots.
• If answer choices are numeric, backsolve by checking both equations.

Professor's tip: Always reject answers that do not make sense in context. Negative time, negative length, or impossible counts should be removed.

Common trap: Reporting both algebraic solutions in a word problem when only one is meaningful.

Session Goals

You will define variables with units, build two competing models, solve their intersections, and interpret which algebraic roots survive physical or practical constraints.

Modeling Requires a Domain

A rectangular garden has area (96) square meters. Its length is (4) meters more than its width. Let width be (w):
\[ A=w(w+4)=96. \]
Then (w^2+4w-96=0), so ((w+12)(w-8)=0). Algebra gives (w=-12) or (w=8), but a geometric length must be positive. The garden is (8) by (12), not (-12) by (-8).

Comparing Two Models

Suppose a company's revenue and cost, in thousands of dollars, are
\[ R(x)=-x^2+18x,qquad C(x)=6x+32. \]
Break-even occurs when (R=C):
\[ -x^2+12x-32=0, \]
or (x^2-12x+32=0). Thus (x=4) or (x=8). Both can be meaningful: profit may be positive between two break-even production levels.

Interpretation Discipline

State the variable, unit, and domain before solving. Afterward, check whether time is nonnegative, lengths are positive, counts are whole numbers when required, and values fall inside the model's stated interval.

Advanced Checkpoint

A projectile has height (h=-2t^2+20t+3), while a rising platform has height (h=4t+35). Setting them equal gives (t^2-8t+16=0), so (t=4). The repeated root means the projectile just touches the moving platform's height at one instant.

Ten Worked Examples

Example 1

Question: A rectangle has area 48 and length 2 more than width. Find its dimensions.

Solution: Let width (x>0). Then (x(x+2)=48), so ((x+8)(x-6)=0). Reject (-8); dimensions are (6) by (8).

Example 2

Question: A projectile has (h=-t^2+6t+4). When is (h=12)?

Solution: Solve (-t^2+6t+4=12), or ((t-2)(t-4)=0). Thus (t=2) and (t=4).

Example 3

Question: A square's area equals four times its side plus 12. Find the positive side.

Solution: Let side be (s>0). Then (s^2=4s+12), so ((s-6)(s+2)=0). The side is (6).

Example 4

Question: Revenue is (R=-x^2+18x), cost is (C=6x+32). Find break-even levels.

Solution: Set (R=C): (x^2-12x+32=0=(x-4)(x-8)). Break-even occurs at (x=4,8).

Example 5

Question: For which production levels is that revenue greater than cost?

Solution: The profit is (-(x-4)(x-8)), positive between its roots. Thus (4<x<8).

Example 6

Question: A ball has (h=-2t^2+20t+3); a platform has (h=4t+35). When do they meet?

Solution: Equate and simplify: (t^2-8t+16=(t-4)^2=0). They meet at (t=4), a tangent-time event.

Example 7

Question: Two positive consecutive integers have product 72. Find them.

Solution: Let the smaller be (n). Then (n(n+1)=72), so ((n-8)(n+9)=0). Positivity gives (n=8); the integers are 8 and 9.

Example 8

Question: A garden has perimeter 28 and area 48. Find its dimensions.

Solution: Let length (x), width (14-x). Then (x(14-x)=48), so (x^2-14x+48=0). Thus (x=6,8); dimensions are 6 by 8.

Example 9

Question: A model yields times (t=-2) and (t=5). Which is physically valid after launch?

Solution: Time after launch requires (t\ge0). Reject (t=-2); the meaningful solution is (t=5).

Example 10

Question: The area of a rectangle is 180; length is 3 times width minus 3. Find width.

Solution: Let width (w>0): (w(3w-3)=180), so (w^2-w-60=0). This does not factor over the integers. The quadratic formula gives (w=(1+\sqrt{241})/2); reject the negative root.