14/14 14. Algebraic Modeling and Applications

A practical guide to translating real-world situations into equations, functions, systems, and inequalities, then solving and interpreting the results with appropriate units and realistic restrictions.

Build and Check a Mathematical Model

A real situation becomes mathematically useful when its quantities and relationships are made explicit. begins by identifying what is known and what must be found. Define variables, include units, organize the information with a diagram or list, and translate the relationships into an equation, inequality, function, or system.

A reliable workflow is:

Read for meaning and identify the unknown quantity.
Define each variable clearly.
Organize the information with a diagram, formula, or graph.
Write the mathematical model.
Solve the model using an appropriate method.
Check the result in the original relationships.
State and interpret the answer in a complete sentence.

Keep units consistent. For example, multiplying $60$ miles per hour by $2$ hours gives $120$ miles. If the units of a result do not match the question, revisit the model or calculation.

Translating a fixed-cost relationship

A gym charges a one-time registration fee of $\$25$ and $\$18$ per month. If $m$ is the number of months and $C$ is the total cost, then

C=25+18m.

The constant term represents the initial fee, and the coefficient of $m$ represents the monthly charge. With a budget of at most $\$151$ , the condition becomes

25+18m\leq151.

Solving gives $m\leq7$ . Thus, the budget covers at most $7$ complete months. The algebra permits real values, but the context restricts the number of complete months to nonnegative integers.

Takeaway: Define variables before calculating, translate every condition, and preserve units and practical restrictions throughout the process.

Model Mixtures and Investments

A combines materials whose concentrations, prices, or rates differ. The key principle is that each contribution equals its rate or concentration multiplied by its quantity:

\text{total contribution}=\text{rate}\times\text{quantity}.

For two materials, the quantities must add to the total quantity, and their contributions must add to the total contribution. If the total amount is $T$ , one quantity can be represented by $x$ and the other by $T-x$ .

Concentration example

To make $20$ liters of a $32\%$ solution from $20\%$ and $50\%$ solutions, let $x$ be the liters of the $20\%$ solution. The amount of pure substance is

0.20x+0.50(20-x)=0.32(20).

Solving gives

0.20x+10-0.50x=6.40,

-0.30x=-3.60,

x=12.

The remaining quantity is $20-12=8$ liters. Therefore, mix $12$ liters of the $20\%$ solution with $8$ liters of the $50\%$ solution. Both quantities are nonnegative and total $20$ liters.

Investment example

Suppose $\$12{,}000$ is divided between an account earning $4\%$ and one earning $9\%$ , with a goal of $\$900$ interest in one year. Let $x$ and $y$ be the amounts invested in the two accounts:

x+y=12{,}000,

0.04x+0.09y=900.

Substituting $x=12{,}000-y$ gives

0.04(12{,}000-y)+0.09y=900,

480+0.05y=900,

y=8{,}400.

Therefore, $x=3{,}600$ . The investment is $\$3{,}600$ at $4\%$ and $\$8{,}400$ at $9\%$ . Checking gives $0.04(3{,}600)+0.09(8{,}400)=900$ .

Takeaway: For mixtures and investments, separate the total-quantity relationship from the total-contribution relationship, then check both conditions.

Use Rate, Motion, and Work Models

For , use

D=rt,

where $D$ is distance, $r$ is rate, and $t$ is time. A list with one entry for each object can help identify the correct distance relationship.

If objects travel to the same destination, their distances may be equal.
If objects move toward each other, their distances add to the initial separation.
If one object catches another, the faster object's distance exceeds the slower object's distance by the initial lead.

Meeting problem

Two cyclists are $180$ miles apart and ride toward each other at $55$ miles per hour and $65$ miles per hour. If $t$ is the time in hours, then

55t+65t=180.

Thus, $120t=180$ , so $t=1.5$ hours. They meet after $1$ hour and $30$ minutes. Their distances, $82.5$ miles and $97.5$ miles, add to $180$ miles.

Catch-up problem

A van is $30$ miles ahead of a car. The van travels at $60$ miles per hour and the car at $75$ miles per hour. After $t$ hours, the car has traveled $30$ miles farther than the van:

75t=60t+30.

Therefore, $15t=30$ , so $t=2$ hours. At that time, the car has traveled $150$ miles and the van $120$ miles.

Work rates

Work problems use rates of job completion, not the number of hours directly. If one worker completes a job in $a$ hours, the worker's rate is

\frac{1}{a}\text{ job per hour}.

Two workers who work together have a combined rate equal to the sum of their rates. If their completion times are $a$ and $b$ , and their combined time is $t$ , then

\frac{1}{a}+\frac{1}{b}=\frac{1}{t}.

For machines that finish a job in $6$ hours and $10$ hours, respectively,

\frac{1}{t}=\frac{1}{6}+\frac{1}{10}=\frac{4}{15},

so $t=\frac{15}{4}=3.75$ hours, or $3$ hours and $45$ minutes. The combined time is less than either individual completion time.

Takeaway: Identify whether distances add, match, or differ by a lead. For work, add completion rates rather than completion times.

Apply Financial Growth Models

Financial models distinguish between interest earned only on the original principal and interest that is repeatedly added to the balance.

For ,

I=Prt,

where $I$ is interest, $P$ is principal, $r$ is the annual rate as a decimal, and $t$ is time in years. The final amount is

A=P+I=P(1+rt).

For $\$2{,}000$ at $4.5\%$ for $3$ years, use $r=0.045$ :

I=(2000)(0.045)(3)=270,

A=2000+270=2270.

The interest is $\$270$ and the final amount is $\$2{,}270$ .

is modeled, when interest is compounded $n$ times per year, by

A=P\left(1+\frac{r}{n}\right)^{nt}.

For continuous compounding, use

A=Pe^{rt}.

For $\$5{,}000$ invested for $2$ years at $6\%$ compounded monthly, $P=5000$ , $r=0.06$ , $n=12$ , and $t=2$ :

A=5000\left(1+\frac{0.06}{12}\right)^{12(2)} =5000(1.005)^{24} \approx5635.80.

The balance is approximately $\$5{,}635.80$ .

Takeaway: Convert percentages to decimals, use years for $t$ in the standard formulas, and distinguish one-time interest calculations from repeated compounding.

Connect Geometry and Motion to Algebra

Geometric relationships become algebraic when a measurement is expressed in terms of an unknown. Draw and label a diagram before substituting into a formula. Common formulas include

\begin{aligned} \text{rectangle area: }&A=lw,\\ \text{rectangle perimeter: }&P=2l+2w,\\ \text{triangle area: }&A=\frac{1}{2}bh,\\ \text{circle area: }&A=\pi r^2,\\ \text{circle circumference: }&C=2\pi r. \end{aligned}

Rectangle example

A rectangle has perimeter $54$ feet, and its length is $3$ feet greater than its width. Let $w$ be the width, so the length is $w+3$ . Substituting into the perimeter formula gives

2(w+3)+2w=54.

Solving gives $w=12$ . The length is $w+3=15$ feet. The dimensions are $12$ feet by $15$ feet, and the check $2(12)+2(15)=54$ confirms the result.

Motion as a

A describes location at time $t$ . For constant velocity,

s(t)=s_0+vt,

where $s_0$ is initial position and $v$ is velocity. For vertical motion near Earth in customary units, a common model is

s(t)=-16t^2+v_0t+s_0.

If a ball is thrown from a height of $20$ feet with initial velocity $48$ feet per second, then

s(t)=-16t^2+48t+20.

To find when it reaches the ground, set $s(t)=0$ :

-16t^2+48t+20=0.

The solutions are approximately $t=2.37$ and $t=-0.37$ . The negative value is rejected because it does not represent a time after the throw. The ball reaches the ground approximately $2.37$ seconds after it is thrown.

Takeaway: Translate geometric or motion descriptions into formulas, then reject algebraic solutions that do not fit the physical situation.

Find Maximum and Minimum Values

seeks the greatest or least possible value of a quantity. The quantity being optimized is represented by an objective function, and the situation determines its realistic .

A typical process is:

Define the decision variable.
Determine the realistic .
Write the objective function.
Find the vertex, critical point, or endpoint values.
Interpret the optimum in context.

For a quadratic function

f(x)=ax^2+bx+c,

the vertex occurs at

x=-\frac{b}{2a}.

If $a<0$ , the vertex gives a maximum; if $a>0$ , it gives a minimum. A restricted may require checking endpoints as well.

Revenue example

A theater sells $120$ tickets at $\$20$ each. For every $\$2$ increase, $10$ fewer tickets are sold. Let $x$ be the number of price increases:

\text{price}=20+2x,

\text{number sold}=120-10x.

Revenue is

R(x)=(20+2x)(120-10x) =-20x^2+40x+2400.

Since the quadratic coefficient is negative, the graph opens downward. Its vertex occurs at

x=-\frac{40}{2(-20)}=1.

The price is $20+2(1)=22$ dollars, and the number sold is $120-10(1)=110$ . The maximum revenue is

R(1)=22(110)=2420.

The model predicts maximum revenue of $\$2{,}420$ at a ticket price of $\$22$ . The realistic is $0\leq x\leq12$ , because the number of tickets sold cannot be negative.

Takeaway: An algebraic optimum is meaningful only after the objective function and its realistic have been identified.

Validate Solutions and Choose Models

A solution is acceptable only if it satisfies the original situation, not merely a transformed equation. Use three checks.

Check the

The of a model must be considered when selecting a solution. Reject values that violate the context. Time, length, area, volume, and mass cannot be negative. Concentrations generally lie between $0\%$ and $100\%$ . Counts of people or objects are usually nonnegative integers. Denominators cannot equal zero, and prices or rates may have additional restrictions.

Check the units

Units should agree with the requested quantity. For example,

(60\text{ miles per hour})(2\text{ hours})=120\text{ miles}.

A result in miles per hour would not answer a question asking for distance.

Check the original conditions

Substitute a proposed answer into the original equation or relationships. This is especially important after multiplying by variables, squaring both sides, or clearing denominators, because such operations can introduce extraneous solutions.

For example, if a geometry equation produces

w=8\quad\text{or}\quad w=-8,

only $w=8$ can represent a rectangle's width. The meaningful conclusion is that the width is $8$ units; the negative root is not physically possible.

Exact and approximate values

An exact answer such as

\frac{15}{4}\text{ hours}

may be preferable to the decimal form $3.75$ hours. When an approximation is required, state the rounding rule and keep sufficient precision during intermediate calculations.

Choosing among common models

A fixed fee plus a constant rate usually uses a linear function, such as $C=a+bx$ .
Two unknown mixture or investment amounts usually use a system of equations.
Constant-speed travel uses $D=rt$ or position functions.
Workers or machines together use a reciprocal-rate equation.
uses $I=Prt$ .
Repeated percentage growth or decay uses an exponential function.
Geometric dimensions use a formula with substitutions.
A maximum or minimum quadratic quantity uses a quadratic function and its vertex.
An unknown exponent or growth time may require a logarithmic equation.

Final takeaway: A complete modeling solution defines variables, builds an appropriate model, solves it accurately, and explains why the selected answer is valid in the original context.