Question 3:
A closed cylindrical tank has a height of h metres and a radius of r metres. The tank can be fully filled with 125/4 π m3 of water. [Use π = 3.142 ]
(a) Find that the total surface area, in m2, of the tank in terms of π and r.
[3 marks]
(b) The tank is made of a material that costs RM 720 per m2. Calculate the minimum cost of the material required to make the tank. [4 marks]
Solution:
(a)
$$ \begin{aligned} & V_{\text {cylinder }}=\frac{125}{4} \pi \mathrm{~m}^3 \\ & \pi r^2 h=\frac{125}{4} \pi \\ & r^2 h=\frac{125}{4} \\ & h=\frac{125}{4 r^2} \end{aligned} $$
$$ \begin{aligned} &\text { Total surface area, } A\\ &\begin{aligned} & =2 \pi r^2+2 \pi r h \\ & =2 \pi r^2+2 \pi r\left(\frac{125}{4 r^2}\right) \\ & =2 \pi r^2+\frac{125 \pi}{2 r} \end{aligned} \end{aligned} $$
(b)
$$ \begin{aligned} A & =2 \pi r^2+\frac{125 \pi}{2 r} \\ & =2 \pi r^2+\frac{125}{2} \pi r^{-1} \\ \frac{\mathrm{~d} A}{\mathrm{~d} r} & =4 \pi r+(-1)\left(\frac{125}{2}\right) \pi r^{-1-1} \\ & =4 \pi r-\frac{125}{2} \pi r^{-2} \\ & =4 \pi r-\frac{125 \pi}{2 r^2} \end{aligned} $$
$$ \begin{aligned} \text { When } \frac{\mathrm{d} A}{\mathrm{~d} r} & =0, \\ 4 \pi r-\frac{125 \pi}{2 r^2} & =0 \\ 4 \pi r & =\frac{125 \pi}{2 r^2} \\ 8 \pi r^3 & =125 \pi \\ 8 r^3 & =125 \\ r^3 & =\sqrt[3]{\frac{125}{8}} \\ r & =\frac{5}{2} \end{aligned} $$
$$ \begin{aligned} & \begin{aligned} \frac{\mathrm{d}^2 A}{\mathrm{~d} r^2} & =4 \pi-(-2)\left(\frac{125}{2}\right) \pi r^{-3} \\ & =4 \pi+\frac{125 \pi}{r^3} \end{aligned} \\ & \text { When } r=\frac{5}{2} \\ & \begin{aligned} \frac{\mathrm{d}^2 A}{\mathrm{~d} r^2} & =4 \pi+\frac{125 \pi}{\left(\frac{5}{2}\right)^3} \\ & =12 \pi>0 \Rightarrow \text { minimum } \end{aligned} \end{aligned} $$
$$ \begin{aligned} A & =2 \pi\left(\frac{5}{2}\right)^2+\frac{125 \pi}{2\left(\frac{5}{2}\right)} \\ & =\frac{25}{2} \pi+25 \pi \\ & =37.5 \pi \mathrm{~m}^2 \\ & =117.825 \mathrm{~m}^2 \end{aligned} $$
$$ \begin{aligned} & \text { Minimum cost } \\ & =117.825 \mathrm{~m}^2 \times \text { RM720 } \\ & =\text { RM84 } 834 \end{aligned} $$
A closed cylindrical tank has a height of h metres and a radius of r metres. The tank can be fully filled with 125/4 π m3 of water. [Use π = 3.142 ]
(a) Find that the total surface area, in m2, of the tank in terms of π and r.
[3 marks]
(b) The tank is made of a material that costs RM 720 per m2. Calculate the minimum cost of the material required to make the tank. [4 marks]
Solution:
(a)
$$ \begin{aligned} & V_{\text {cylinder }}=\frac{125}{4} \pi \mathrm{~m}^3 \\ & \pi r^2 h=\frac{125}{4} \pi \\ & r^2 h=\frac{125}{4} \\ & h=\frac{125}{4 r^2} \end{aligned} $$
$$ \begin{aligned} &\text { Total surface area, } A\\ &\begin{aligned} & =2 \pi r^2+2 \pi r h \\ & =2 \pi r^2+2 \pi r\left(\frac{125}{4 r^2}\right) \\ & =2 \pi r^2+\frac{125 \pi}{2 r} \end{aligned} \end{aligned} $$
(b)
$$ \begin{aligned} A & =2 \pi r^2+\frac{125 \pi}{2 r} \\ & =2 \pi r^2+\frac{125}{2} \pi r^{-1} \\ \frac{\mathrm{~d} A}{\mathrm{~d} r} & =4 \pi r+(-1)\left(\frac{125}{2}\right) \pi r^{-1-1} \\ & =4 \pi r-\frac{125}{2} \pi r^{-2} \\ & =4 \pi r-\frac{125 \pi}{2 r^2} \end{aligned} $$
$$ \begin{aligned} \text { When } \frac{\mathrm{d} A}{\mathrm{~d} r} & =0, \\ 4 \pi r-\frac{125 \pi}{2 r^2} & =0 \\ 4 \pi r & =\frac{125 \pi}{2 r^2} \\ 8 \pi r^3 & =125 \pi \\ 8 r^3 & =125 \\ r^3 & =\sqrt[3]{\frac{125}{8}} \\ r & =\frac{5}{2} \end{aligned} $$
$$ \begin{aligned} & \begin{aligned} \frac{\mathrm{d}^2 A}{\mathrm{~d} r^2} & =4 \pi-(-2)\left(\frac{125}{2}\right) \pi r^{-3} \\ & =4 \pi+\frac{125 \pi}{r^3} \end{aligned} \\ & \text { When } r=\frac{5}{2} \\ & \begin{aligned} \frac{\mathrm{d}^2 A}{\mathrm{~d} r^2} & =4 \pi+\frac{125 \pi}{\left(\frac{5}{2}\right)^3} \\ & =12 \pi>0 \Rightarrow \text { minimum } \end{aligned} \end{aligned} $$
$$ \begin{aligned} A & =2 \pi\left(\frac{5}{2}\right)^2+\frac{125 \pi}{2\left(\frac{5}{2}\right)} \\ & =\frac{25}{2} \pi+25 \pi \\ & =37.5 \pi \mathrm{~m}^2 \\ & =117.825 \mathrm{~m}^2 \end{aligned} $$
$$ \begin{aligned} & \text { Minimum cost } \\ & =117.825 \mathrm{~m}^2 \times \text { RM720 } \\ & =\text { RM84 } 834 \end{aligned} $$