Polynomial word problems are solved by translating a real situation into a polynomial model.
After the model is built, ordinary algebra tools such as expansion, factoring, roots, vertices, and optimization
can be used to answer the real-world question.
1. General method for polynomial word problems
A reliable process is:
- Define the variable clearly.
- Write the constraint or relationship from the problem.
- Build a polynomial or polynomial-based model.
- Solve the algebraic question: evaluate, maximize, minimize, or find roots.
- Interpret the answer in the practical domain.
The practical domain matters. For example, a length cannot be negative, and time after launch usually means
\(t\geq 0\).
2. Geometry optimization: box with square base
Suppose a box has a square base of side \(x\), height \(h\), and fixed volume \(V\).
The volume equation is:
\[
\begin{aligned}
V=x^2h.
\end{aligned}
\]
Therefore:
\[
\begin{aligned}
h=\frac{V}{x^2}.
\end{aligned}
\]
For a closed box, the surface area is:
\[
\begin{aligned}
S=2x^2+4xh.
\end{aligned}
\]
Substitute \(h=V/x^2\):
\[
\begin{aligned}
S(x)
&=
2x^2+4x\left(\frac{V}{x^2}\right)\\
&=
2x^2+\frac{4V}{x}.
\end{aligned}
\]
Minimizing this model gives the most efficient dimensions. For a closed square-base box, the optimal box is a cube:
\[
\begin{aligned}
x=h=\sqrt[3]{V}.
\end{aligned}
\]
3. Geometry optimization: rectangle with fixed perimeter
If a rectangle has fixed perimeter \(P\), width \(x\), and length \(L\), then:
\[
\begin{aligned}
2x+2L=P.
\end{aligned}
\]
Solve for length:
\[
\begin{aligned}
L=\frac{P}{2}-x.
\end{aligned}
\]
The area model is:
\[
\begin{aligned}
A(x)
&=
x\left(\frac{P}{2}-x\right)\\
&=
-x^2+\frac{P}{2}x.
\end{aligned}
\]
This is a downward-opening quadratic, so its vertex gives the maximum area.
The maximum happens when:
\[
\begin{aligned}
x=\frac{P}{4}.
\end{aligned}
\]
The rectangle with maximum area for a fixed perimeter is a square.
4. Physics: projectile height model
A common physics polynomial model is vertical projectile motion:
\[
\begin{aligned}
h(t)=h_0+v_0t-\frac{1}{2}gt^2.
\end{aligned}
\]
This is a quadratic in time. Since the coefficient of \(t^2\) is negative, the graph opens downward.
The vertex gives the maximum height:
\[
\begin{aligned}
t_{\max}=\frac{v_0}{g}.
\end{aligned}
\]
To find when the projectile hits the ground, solve:
\[
\begin{aligned}
h(t)=0.
\end{aligned}
\]
Only nonnegative time values make sense in the physical situation.
5. Business: revenue and profit
A linear demand model often has the form:
\[
\begin{aligned}
p(q)=a-bq,
\end{aligned}
\]
where \(q\) is quantity and \(p(q)\) is price.
Revenue is:
\[
\begin{aligned}
R(q)=q\cdot p(q)=q(a-bq)=aq-bq^2.
\end{aligned}
\]
If variable cost per unit is \(c\), and fixed cost is \(F\), then profit is:
\[
\begin{aligned}
P(q)
&=
R(q)-cq-F\\
&=
-bq^2+(a-c)q-F.
\end{aligned}
\]
This is a downward-opening quadratic, so maximum profit occurs at the vertex:
\[
\begin{aligned}
q=\frac{a-c}{2b}.
\end{aligned}
\]
6. Roots in applications
Roots often represent important real-world thresholds.
- In projectile motion, roots of \(h(t)=0\) represent ground contact times.
- In profit models, roots of \(P(q)=0\) represent break-even quantities.
- In geometry, roots may represent impossible or boundary dimensions.
Not every algebraic root is meaningful. Always check the practical domain.
7. Optimization with quadratics
For a quadratic:
\[
\begin{aligned}
f(x)=ax^2+bx+c,
\end{aligned}
\]
the vertex occurs at:
\[
\begin{aligned}
x=-\frac{b}{2a}.
\end{aligned}
\]
If \(a>0\), the vertex is a minimum. If \(a<0\), the vertex is a maximum.
8. Optimization with derivatives
For advanced problems, the derivative gives the rate of change.
Maximum or minimum points usually occur where:
\[
\begin{aligned}
f'(x)=0.
\end{aligned}
\]
Example: for the closed square-base box surface area
\[
\begin{aligned}
S(x)=2x^2+\frac{4V}{x},
\end{aligned}
\]
the derivative is:
\[
\begin{aligned}
S'(x)=4x-\frac{4V}{x^2}.
\end{aligned}
\]
Setting \(S'(x)=0\) gives:
\[
\begin{aligned}
x^3=V.
\end{aligned}
\]
9. Formula summary
The table below uses plain text formulas in table cells to avoid raw LaTeX rendering problems.
10. Common mistakes
- Using an algebraic answer that does not make sense in the practical domain.
- Forgetting that dimensions, time, quantity, and volume cannot usually be negative.
- Maximizing the wrong expression, such as maximizing perimeter instead of area.
- Forgetting fixed costs in profit problems.
- Using both roots of a quadratic when only one root is physically meaningful.
- Not defining the variable before building the polynomial model.
Key idea: polynomial word problems are solved by modeling first, solving second, and interpreting last.
