一家工厂为公司活动打印海报。不同尺寸的海报打印价格如下。该工厂对打印机的操作有以下指南。

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.