2.6 Modeling with Functions

2.6.1 Learning Objectives

By the end of this section, you should be able to:

define variables and translate relationships into a function model;
determine a meaningful domain for a model;
construct geometry, revenue, and profit functions;
optimize a model algebraically or graphically; and
interpret a model’s output using appropriate units and constraints.

2.6.2 Building a Function Model

A mathematical model uses mathematics to represent a real situation. A useful process is to define all variables with units, write the relationships among them, express the target quantity as a function of one variable, state the meaningful domain, analyze the function, and interpret the result.

Example 2.28 Modeling a Rectangle

A rectangle has perimeter $40\text{ m}$. Write its area as a function of its width $w$.

Solution

Let $L$ be the length in metres. From $2L+2w=40$, $L=20-w$. Therefore,

\[ \boxed{A(w)=w(20-w),\quad 0<w<20}. \]

Exercise 2.27 A rectangle has perimeter $60\text{ cm}$. Write its area as a function of its width $w$.

2.6.3 Geometry Models

Geometry models often begin with a formula for area, volume, distance, or perimeter. Use the condition in the problem to eliminate all but one independent variable.

Example 2.29 Modeling an Open Box

Squares of side $x$ centimetres are cut from each corner of a $30\text{ cm}$ by $20\text{ cm}$ sheet. Write the volume of the resulting open box.

Solution

The box has height $x$, length $30-2x$, and width $20-2x$. Hence,

\[ \boxed{V(x)=x(30-2x)(20-2x),\quad 0<x<10}. \]

Exercise 2.28 Write the open-box volume function for a $24\text{ cm}$ by $16\text{ cm}$ sheet.

2.6.4 Revenue, Cost, and Profit

If $x$ items are sold at price $p(x)$, revenue is $R(x)=xp(x)$. If $C(x)$ is the cost, profit is

\[ P(x)=R(x)-C(x). \]

In practice, the number of items may need to be a non-negative integer.

Example 2.30 Maximizing Ticket Revenue

A theatre sells $300$ tickets at $\$20$. For each $\$1$ increase, $10$ fewer tickets are sold. Find the price that maximizes predicted revenue.

Solution

Let $x$ be the number of one-dollar increases. Then

\[ R(x)=(20+x)(300-10x)=-10x^2+100x+6000. \]

The vertex occurs at $x=-100/[2(-10)]=5$. The price is therefore $\$25$, and

\[ R(5)=25(250)=6250. \]

\[ \boxed{\text{Price }\$25;\quad \text{maximum predicted revenue }\$6250} \]

Exercise 2.29 A fundraiser sells $200$ tickets at $\$12$. For each $\$1$ increase, $5$ fewer tickets are sold. Write the revenue function.

2.6.5 Optimizing a Model

To optimize a model is to find its greatest or least useful value. A quadratic model can be optimized at its vertex. For other models, Desmos can locate an approximate extreme value on a restricted domain. Always check practical constraints and permitted inputs.

Example 2.31 Maximizing an Enclosed Area

A farmer has $200\text{ m}$ of fencing for three sides of a rectangle beside a river. Find the dimensions that maximize the area.

Solution

Let $x$ be each side perpendicular to the river. The third fenced side is $200-2x$, so

\[ A(x)=x(200-2x)=-2x^2+200x. \]

The vertex occurs at $x=50$, making the third side $100$. Thus,

\[ \boxed{50\text{ m by }100\text{ m};\quad \text{maximum area }5000\text{ m}^2}. \]

Exercise 2.30 A rectangle has perimeter $40\text{ m}$. Find the dimensions that maximize its area.

2.6.6 Graphical and Numerical Optimization

In Desmos, enter the model with its domain restriction, select the relevant turning point, and interpret both coordinates with units. The context determines appropriate rounding.

Example 2.32 Maximizing Open-Box Volume

Use Desmos to maximize

\[ V(x)=x(30-2x)(20-2x),\qquad 0<x<10. \]

Solution

The relevant turning point occurs near $x=3.92$, where $V(x)\approx1056.3$.

\[ \boxed{\text{Maximum volume }\approx1056.3\text{ cm}^3} \]

Exercise 2.31 Use Desmos to maximize $V(x)=x(24-2x)(16-2x)$ on $0<x<8$.

2.6.7 Conceptual Takeaways

A model links mathematical variables to quantities in a real situation.
Definitions, units, and constraints determine what a formula means.
A meaningful domain can be narrower than a function’s algebraic domain.
An algebraic optimum is useful only if it is possible in context.
Numerical results should be rounded appropriately for the situation.

2.6.8 Skills You Should Be Able to Do

Define variables with units and translate constraints into equations.
Express a target quantity as a function of one variable.
State and justify a model’s meaningful domain.
Construct geometry, revenue, cost, and profit models.
Optimize a model and interpret the result.

2.6.9 Practice Problems with Solutions

A rectangle has perimeter $52\text{ cm}$. Write its area as a function of width $w$.

Show Solution

Since $L=26-w$,

\[ \boxed{A(w)=w(26-w),\quad 0<w<26}. \]
A square has area $A$. Write its perimeter as a function of $A$.

Show Solution

Since its side is $\sqrt A$,

\[ \boxed{P(A)=4\sqrt A,\quad A\ge0}. \]
A $20\text{ cm}$ by $14\text{ cm}$ sheet forms an open box after squares of side $x$ are removed. Write the volume function and domain.

Show Solution

\[ \boxed{V(x)=x(20-2x)(14-2x),\quad 0<x<7}. \]
A company sells $x$ items at $p(x)=80-0.5x$ dollars each. Write its revenue function.

Show Solution

\[ \boxed{R(x)=x(80-0.5x)=-0.5x^2+80x}. \]
For Problem 4, find the maximum revenue.

Show Solution

The vertex occurs at $x=-80/[2(-0.5)]=80$. Then $R(80)=3200$, so

\[ \boxed{80\text{ items};\quad \$3200}. \]
A school sells $240$ tickets at $\$15$. Each $\$1$ increase reduces sales by $8$ tickets. What whole-dollar price gives the greatest predicted revenue?

Show Solution

\[ R(x)=(15+x)(240-8x)=-8x^2+120x+3600. \]

The vertex is at $x=7.5$. Testing $x=7$ and $x=8$ gives

\[ R(7)=4048,\qquad R(8)=4048. \]

Thus, $\boxed{\$22\text{ or }\$23}$ gives predicted revenue $\$4048$.
A farmer has $180\text{ m}$ of fencing for three sides of a rectangle beside a barn. Find the dimensions that maximize area.

Show Solution

\[ A(x)=x(180-2x)=-2x^2+180x. \]

Its vertex is at $x=45$, so the other dimension is $90$.

\[ \boxed{45\text{ m by }90\text{ m};\quad 4050\text{ m}^2}. \]
A closed cylinder has volume $500\text{ cm}^3$. Express its surface area as a function of radius $r$.

Show Solution

From $\pi r^2h=500$, $h=500/(\pi r^2)$. Therefore,

\[ S(r)=2\pi r^2+2\pi r\left(\frac{500}{\pi r^2}\right). \]

Hence,

\[ \boxed{S(r)=2\pi r^2+\frac{1000}{r},\quad r>0}. \]