Polynomial modeling is the process of translating a real-world situation into a polynomial function.
In many applications, the goal is not just to evaluate the polynomial, but to optimize it. For example,
you may want the largest possible volume, the largest possible area, or the greatest possible profit.
1. General optimization strategy
Most polynomial optimization problems follow the same structure. First, choose a variable. Then write
the quantity you want to maximize or minimize as a function of that variable. Finally, use calculus or
vertex methods to locate the best feasible value.
General derivative condition.
\[
\begin{aligned}
f'(x) &= 0
\end{aligned}
\]
A solution of \(f'(x)=0\) is called a critical point. In an optimization problem, you must also check
whether the point is inside the physical domain of the problem.
Second derivative maximum test.
\[
\begin{aligned}
f''(x_0) &< 0
\end{aligned}
\]
If \(f''(x_0)<0\), the graph is concave down at \(x_0\), so the critical point is a local maximum.
In many box, area, and profit models, the feasible domain is a bounded interval and the interior
critical point is the maximum.
2. Open-top box from a rectangular sheet
A common volume problem starts with a rectangular sheet of length \(L\) and width \(W\). Squares of
side \(x\) are cut from each corner, and the sides are folded upward. The height of the box is \(x\),
the base length is \(L-2x\), and the base width is \(W-2x\).
Volume model.
\[
\begin{aligned}
V(x) &= x(L-2x)(W-2x)
\end{aligned}
\]
The physical domain is not all real numbers. The cut size must be positive, and the remaining base
dimensions must also be positive.
Feasible domain.
\[
\begin{aligned}
0 &< x < \frac{\min(L,W)}{2}
\end{aligned}
\]
After expanding, \(V(x)\) is a cubic polynomial. The maximum occurs at the feasible critical point.
Derivative of the open-box model.
\[
\begin{aligned}
V(x) &= x(L-2x)(W-2x) \\
&= 4x^3-2(L+W)x^2+LWx \\
V'(x) &= 12x^2-4(L+W)x+LW
\end{aligned}
\]
3. Square-base box with fixed surface area
Another common model uses a square base of side \(x\) and height \(h\). For an open-top box, the
surface area is base plus four side faces.
Open-top surface-area constraint.
\[
\begin{aligned}
S &= x^2+4xh \\
h &= \frac{S-x^2}{4x}
\end{aligned}
\]
Substitute this height into \(V=x^2h\).
Open-top volume polynomial.
\[
\begin{aligned}
V(x) &= x^2\left(\frac{S-x^2}{4x}\right) \\
&= \frac{Sx-x^3}{4}
\end{aligned}
\]
For a closed box, the surface area includes the top and bottom:
Closed-box surface-area model.
\[
\begin{aligned}
S &= 2x^2+4xh \\
h &= \frac{S-2x^2}{4x} \\
V(x) &= \frac{Sx-2x^3}{4}
\end{aligned}
\]
The open-top maximum occurs when the height is half the base side. The closed-box maximum occurs
when the box is a cube.
4. Fixed total edge length
A closed box with square base has eight horizontal edges of length \(x\) and four vertical edges
of length \(h\). If the total edge length is \(T\), then:
Edge-length constraint.
\[
\begin{aligned}
T &= 8x+4h \\
h &= \frac{T}{4}-2x
\end{aligned}
\]
The volume becomes a cubic polynomial.
Volume from fixed total edge length.
\[
\begin{aligned}
V(x) &= x^2h \\
&= x^2\left(\frac{T}{4}-2x\right) \\
&= \frac{T}{4}x^2-2x^3
\end{aligned}
\]
Differentiating gives:
\[
\begin{aligned}
V'(x) &= \frac{T}{2}x-6x^2 \\
&= x\left(\frac{T}{2}-6x\right)
\end{aligned}
\]
The nonzero critical point is \(x=T/12\), and then \(h=T/12\). So the maximum-volume box is a cube.
5. Rectangle with fixed perimeter
If a rectangle has perimeter \(P\), width \(x\), and length \(\ell\), then:
Perimeter constraint.
\[
\begin{aligned}
2x+2\ell &= P \\
\ell &= \frac{P}{2}-x
\end{aligned}
\]
The area model is:
Area polynomial.
\[
\begin{aligned}
A(x) &= x\left(\frac{P}{2}-x\right) \\
&= \frac{P}{2}x-x^2
\end{aligned}
\]
This is a downward-opening quadratic. Its maximum occurs at the vertex:
\[
\begin{aligned}
x_{\max} &= \frac{P}{4}
\end{aligned}
\]
Therefore, among all rectangles with a fixed perimeter, the square has the greatest area.
6. Fencing a garden against one wall
If one side of a rectangular garden is a wall, only three sides need fencing. Let \(x\) be each of
the two equal side widths, and let \(\ell\) be the third fenced side.
Fencing constraint and area model.
\[
\begin{aligned}
2x+\ell &= F \\
\ell &= F-2x \\
A(x) &= x(F-2x) \\
&= Fx-2x^2
\end{aligned}
\]
The maximum occurs when:
\[
\begin{aligned}
A'(x) &= F-4x \\
x_{\max} &= \frac{F}{4}
\end{aligned}
\]
The fenced length is then \(F/2\), which is twice the side width.
7. Polynomial profit models
In business applications, a simple linear demand model often has the form:
Linear demand.
\[
\begin{aligned}
p(q) &= a-bq
\end{aligned}
\]
Here \(q\) is quantity, \(a\) is the price intercept, and \(b\) is the demand slope. Revenue is
price times quantity, while cost may be modeled as \(C(q)=cq+F\).
Profit polynomial.
\[
\begin{aligned}
\Pi(q) &= R(q)-C(q) \\
&= q(a-bq)-(cq+F) \\
&= -bq^2+(a-c)q-F
\end{aligned}
\]
Since \(b>0\), the profit polynomial opens downward. The maximizing quantity is:
\[
\begin{aligned}
q_{\max} &= \frac{a-c}{2b}
\end{aligned}
\]
This value must still be checked against the feasible domain. For example, quantity cannot be
negative, and price usually should not be negative.
8. Integer constraints
Some real-world problems require whole-number dimensions or whole-number production quantities.
In that case, the continuous maximum may not be usable. A good strategy is to find the continuous
maximum first and then test nearby integers.
Integer check idea.
\[
\begin{aligned}
x_{\text{integer}} &\in \{\lfloor x_{\max}\rfloor,\ \lceil x_{\max}\rceil\}
\end{aligned}
\]
The best integer answer is the feasible integer near the continuous optimum that gives the largest
model output.
9. Common modeling mistakes
- Do not optimize before defining the feasible domain.
- Do not forget physical restrictions such as positive length, positive height, or nonnegative quantity.
- Do not confuse the variable \(x\) with the final dimensions; usually other dimensions depend on \(x\).
- Do not assume the critical point is feasible; check the interval.
- Do not round too early. Compute with full precision and round only the final result.
10. Final checklist
- Choose the variable.
- Write constraints using the given information.
- Substitute constraints into the quantity being optimized.
- Simplify to a polynomial model.
- Find the derivative or use the quadratic vertex formula.
- Check the feasible domain.
- Evaluate the maximum or minimum.
- Translate the result back into real-world dimensions or business quantities.
