Question 10:
Diagram 7 shows a roll of aluminium.
The aluminium roll has a length of 55440 mm and can be cut exactly into 72 sections. The length of the first sheet is x mm and the length of each subsequent sheet increased by 20 mm. Atan is assigned to cut the aluminium roll.
(a) State the length, in mm, of the first three aluminium sheets in terms of x. [1 mark]
(b) Atan has completed 2/3 of his task.
By determining the value of x, calculate the total length, in mm , of the last 18 sheets that have been cut. [ 4 marks]
(c) The cuts of aluminium sheets can be formed into an open cylinder at both ends in two different ways. When Atan completed his task, his brother has chosen the last sheet to form a cylinder with maximum volume.
By calculation, suggest the cylinder that fulfils the condition. [3 marks]
Answer:
(a) $$ (x, x+20, x+40) \mathrm{mm} $$
(b) $$ \begin{aligned} & a=x \\ & T_{72}=x+(72-1)(20) \\ & =x+1420 \end{aligned} $$
$$ \begin{aligned} S_{72} & =55440 \\ \frac{72}{2}(x+x+1420) & =55440 \\ 2 x+1420 & =\frac{55440}{36} \\ x & =60 \end{aligned} $$
$$ \text { Last } 18 \text { terms: } T_{55}, T_{56}, \ldots, T_{72} $$
$$ \begin{aligned} S_{18} & =\frac{18}{2}[2(60+30(20))+(18-1)(20)] \\ & =14940 \mathrm{~mm} \end{aligned} $$
(c) Method 1
$$ 60+(72-1)(20)=1480 $$
$$ \begin{aligned} 2 \pi r & =1480 \\ r & =\frac{1480}{2} \\ & =\frac{740}{\pi} \end{aligned} $$
$$ 60+(72-1)(20)=1480 \text { and } r=\frac{740}{\pi} $$
$$ \begin{aligned} V_1 & =\pi r^2 h \\ & =\pi\left(\frac{740}{\pi}\right)^2(500) \\ & =\frac{273800000}{\pi} \mathrm{~mm}^3 \\ & =\frac{2.738 \times 10^8}{\pi} \mathrm{~mm}^3 \end{aligned} $$
Method 2:
$$
\begin{aligned}
2 \pi r & =500 \\
r & =\frac{500}{2 \pi} \\
& =\frac{250}{\pi}
\end{aligned}
$$
$$ \begin{aligned} V_2 & =\pi r^2 h \\ & =\pi\left(\frac{250}{\pi}\right)^2(1480) \\ & =\frac{92500000}{\pi} \mathrm{~mm}^3 \\ & =\frac{9.25 \times 10^7}{\pi} \mathrm{~mm}^3 \end{aligned} $$
$$ V_1=\pi\left(\frac{740}{\pi}\right)^2(500) \text { and } V_2=\pi\left(\frac{250}{\pi}\right)^2(1480) $$
$$ \text { Method } 1 \text { forms a cylinder with maximum volume:: } $$
$$ V_1=\frac{2.738 \times 10^8}{\pi} \mathrm{~mm}^3 \text { and } r=\frac{740}{\pi} $$
Diagram 7 shows a roll of aluminium.
The aluminium roll has a length of 55440 mm and can be cut exactly into 72 sections. The length of the first sheet is x mm and the length of each subsequent sheet increased by 20 mm. Atan is assigned to cut the aluminium roll.(a) State the length, in mm, of the first three aluminium sheets in terms of x. [1 mark]
(b) Atan has completed 2/3 of his task.
By determining the value of x, calculate the total length, in mm , of the last 18 sheets that have been cut. [ 4 marks]
(c) The cuts of aluminium sheets can be formed into an open cylinder at both ends in two different ways. When Atan completed his task, his brother has chosen the last sheet to form a cylinder with maximum volume.
By calculation, suggest the cylinder that fulfils the condition. [3 marks]
Answer:
(a) $$ (x, x+20, x+40) \mathrm{mm} $$
(b) $$ \begin{aligned} & a=x \\ & T_{72}=x+(72-1)(20) \\ & =x+1420 \end{aligned} $$
$$ \begin{aligned} S_{72} & =55440 \\ \frac{72}{2}(x+x+1420) & =55440 \\ 2 x+1420 & =\frac{55440}{36} \\ x & =60 \end{aligned} $$
$$ \text { Last } 18 \text { terms: } T_{55}, T_{56}, \ldots, T_{72} $$
$$ \begin{aligned} S_{18} & =\frac{18}{2}[2(60+30(20))+(18-1)(20)] \\ & =14940 \mathrm{~mm} \end{aligned} $$
(c) Method 1
$$ 60+(72-1)(20)=1480 $$
$$ \begin{aligned} 2 \pi r & =1480 \\ r & =\frac{1480}{2} \\ & =\frac{740}{\pi} \end{aligned} $$
$$ 60+(72-1)(20)=1480 \text { and } r=\frac{740}{\pi} $$
$$ \begin{aligned} V_1 & =\pi r^2 h \\ & =\pi\left(\frac{740}{\pi}\right)^2(500) \\ & =\frac{273800000}{\pi} \mathrm{~mm}^3 \\ & =\frac{2.738 \times 10^8}{\pi} \mathrm{~mm}^3 \end{aligned} $$
Method 2:
$$
\begin{aligned}
2 \pi r & =500 \\
r & =\frac{500}{2 \pi} \\
& =\frac{250}{\pi}
\end{aligned}
$$$$ \begin{aligned} V_2 & =\pi r^2 h \\ & =\pi\left(\frac{250}{\pi}\right)^2(1480) \\ & =\frac{92500000}{\pi} \mathrm{~mm}^3 \\ & =\frac{9.25 \times 10^7}{\pi} \mathrm{~mm}^3 \end{aligned} $$
$$ V_1=\pi\left(\frac{740}{\pi}\right)^2(500) \text { and } V_2=\pi\left(\frac{250}{\pi}\right)^2(1480) $$
$$ \text { Method } 1 \text { forms a cylinder with maximum volume:: } $$
$$ V_1=\frac{2.738 \times 10^8}{\pi} \mathrm{~mm}^3 \text { and } r=\frac{740}{\pi} $$