Linear Models

Section 5.6 Linear Models

Linear equations are frequently used to construct simple models of the interaction between two quantities. We will illustrate this with a few examples below. Think about the appropriate choice of linear form in each example.

Standard Form: $Ax+By = C$
Point-Slope Form: $y-y_0 = m(x-x_0)$
Slope-Intercept Form: $y = mx+b$

Example 5.15.

A rectangular fence is to be constructed from a fixed budget of 1000 dollars. Two parallel sides of the fence are constructed from a material costing 4 dollars per foot, while the remaining sides are constructed from a material costing 3 dollars per foot. Express these constraints as a linear equation.

Solution.

Let $x$ denote the length of the sides costing 4 dollars per foot and let $y$ denote the length of the sides costing 3 dollars per foot. Then the cost of the fence is $4x+4x+3y+3y = 8x + 6y\text{.}$ The fixed budget imposes the linear constraint

\begin{equation*} 8x+6y = 1000. \end{equation*}

Example 5.16.

A gasoline-powered generator has a fuel tank capacity of $10$ gallons and consumes fuel at a rate $1.5$ gallons per hour. Find a linear model for the fuel consumption over time. How long can the generator run on a full tank?

Solution.

Let $y$ be the amount of fuel in the tank after running for $x$ hours. Then initially, $y=10\text{,}$ when $x=0\text{,}$ determining the $y$-intercept. The rate of decrease of fuel provides the slope $-1.5\text{.}$ So,

\begin{equation*} y = -1.5x + 10. \end{equation*}

The tank will be empty when

\begin{equation*} 0 = -1.5x + 10 \end{equation*}

which we solve to find $x=10/1.5 \approx 6.7$ hours.

Example 5.17.

A company earns a revenue of $\$1000$ from selling $25$ units of their product. They earn $\$1650$ from selling $41$ units. Find a linear relationship modeling the revenue in terms of number of units sold.

Solution.

Let $R$ denote the revenue and $x$ denote the number of units sold.

The data provided allows us to find the rate at which revenue changes as we increase the number of units sold:

\begin{equation*} \frac{\Delta R}{\Delta x} = \frac{1650-1000}{41-25} = \frac{650}{16} = 40.625 \end{equation*}

So the slope of our line is $m = 40.625$ dollars per unit. We can now use point-slope form along with either data point to find a linear equation

\begin{align*} R - 1000 \amp= 40.625(x-25)\\ R \amp= 40.625(x-25)+1000\\ R \amp= 40.625(x-25)+1000\\ R \amp= 40.625x-15.625 \end{align*}

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