6.7Real-World Applications of Linear Functions

The gradient is usually the rate of change; the $y$ -intercept is usually the starting value.

The bill against units used. The intercept is the fixed charge; the gradient is the rate per kWh.

Fixed charge + rate × quantity: phone, electricity and hire-charge bills all take this shape.
Distance-time: distance $=$ speed $\times$ time; the gradient is the speed.

Electricity bill

A household pays a fixed charge of $\displaystyle \$8$ per month plus $\displaystyle \$0.25$ per kWh of electricity used. Write an equation for the monthly bill $\displaystyle B$ in terms of units used $\displaystyle u$ . Find the bill when 240 kWh are used.

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B = 8 + 0.25u \quad \scriptsize\textit{fixed charge + rate }\times\textit{ units}

When

u = 240

\displaystyle B

\displaystyle = 8 + 0.25(240)

\displaystyle = 8 + 60

\displaystyle = \$68

The gradient (

0.25

) is the cost per kWh; the

y

-intercept (

8

) is the fixed monthly charge.

Now you try

A taxi charges a flag-down fare of $\$3.40$ and $\$0.40$ per km. Find the total fare for a journey of 8 km.

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\displaystyle \text{Total fare}

\displaystyle = 3.40 + 0.40 \times 8

\displaystyle = 3.40 + 3.20

\displaystyle = \$6.60

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Finding the linear model from two data points

A catering company charges a delivery fee plus a fixed amount per guest. For 30 guests the total cost is $\displaystyle \$410$ , and for 50 guests it is $\displaystyle \$610$ . Find the cost for 80 guests.

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The rate per guest is the gradient of the cost line.

\displaystyle r

\displaystyle = \dfrac{610 - 410}{50 - 30}

\displaystyle = \dfrac{200}{20}

= 10

, so each guest adds

\$10

\displaystyle \text{Delivery fee}

\displaystyle = 410 - 30 \times 10

\displaystyle = \$110

Cost for

80

guests

= 110 + 10(80)

= \$910

Now you try

The temperature in °C can be converted to °F using the formula $F = 1.8C + 32$ . Find the temperature in °C when it is 104°F.

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\displaystyle F

\displaystyle = 1.8C + 32

\displaystyle 104

\displaystyle = 1.8C + 32 \quad \scriptsize\textit{substitute } F = 104

\displaystyle 1.8C

\displaystyle = 104 - 32

\displaystyle 1.8C

\displaystyle = 72

\displaystyle C

\displaystyle = \frac{72}{1.8}

\displaystyle = 40\degree\text{C}

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