A factory prints posters for corporate events. The prices of printing posters of different sizes are listed below. The factory has the following guidelines for operating the printers.

1. Convert the poster dimensions to centimetres and find the area in square centimetres.

Bulletin poster size: $11 \times 17$ inches.

Since $1$ inch $=2.54$ cm,
$11\text{ inches}=11\times2.54=27.94\text{ cm}$
$17\text{ inches}=17\times2.54=43.18\text{ cm}$
Area
$=27.94\times43.18=120.6448\text{ cm}^2$

2. Convert the area from square centimetres to square metres.

Since $1\text{ m}=100\text{ cm}$,
$1\text{ m}^2=10000\text{ cm}^2$
$120.6448\text{ cm}^2=\frac{120.6448}{10000}=0.01206448\text{ m}^2$

1. Identify the price per bulletin poster.

From the table, the price of one bulletin poster is $2.00.

2. Multiply the unit price by the number of bulletin posters.

Cost of $1200$ bulletin posters

$1200 \times 2.00 = 2400$

1. Find the actual printing time available in one day.

The printer operates from 9.00 am to 4.00 pm, which is $7$ hours.

It must cool for $30$ minutes after every $2$ hours of printing.

A possible schedule is:

$9.00$ am to $11.00$ am: print $2$ h

$11.00$ am to $11.30$ am: cool $0.5$ h

$11.30$ am to $1.30$ pm: print $2$ h

$1.30$ pm to $2.00$ pm: cool $0.5$ h

$2.00$ pm to $4.00$ pm: print $2$ h

So the total printing time is

$2+2+2=6\text{ h}$

$6\text{ h}=6\times 60\times 60=21600\text{ s}$

2. Form an equation for the number of posters printed by both printers.

Let the newer printer take $x$ seconds to print one bulletin poster.

Then the older printer takes $x+15$ seconds to print one bulletin poster.

In $21600$ seconds,

number printed by newer printer $=\dfrac{21600}{x}$,

number printed by older printer $=\dfrac{21600}{x+15}$.

Together, they print $1200$ posters, so

$\frac{21600}{x}+\frac{21600}{x+15}=1200$

3. Solve the equation to find the time taken by each printer per poster.

Divide the equation by $1200$:

$\frac{18}{x}+\frac{18}{x+15}=1$

Multiply throughout by $x(x+15)$:

$18(x+15)+18x=x(x+15)$

$18x+270+18x=x^2+15x$

$36x+270=x^2+15x$

$x^2-21x-270=0$

$\left(x-30\right)\left(x+9\right)=0$

Since time cannot be negative,

$x=30$

So the newer printer takes $30$ seconds per poster, and the older printer takes

$30+15=45\text{ seconds per poster}.$

4. Find the total area printed by the older printer in one day.

Number of posters printed by the older printer in one day:

$\frac{21600}{45}=480$

From part (a), area of one bulletin poster is

$0.01206448\text{ m}^2$

Total area printed by the older printer:

$480\times 0.01206448=5.791\,\text{m}^2\text{ approximately}$

5. Compare the older printer's daily printing area with the servicing guideline and conclude.

The printer needs servicing if its rate of printing has slowed down to less than $60\text{ m}^2$ a day.

For the older printer,

$5.791\text{ m}^2<60\text{ m}^2$

Therefore, the older printer needs servicing.