- $C=k_1+k_2A$, where $k_1$ and $k_2$ are non-zero constants.
For $A=2$ and $C=62$, we have
$\begin{array}{ll}
62=k_1+2k_2 & \ldots \unicode{x2460}
\end{array}$For $A=6$ and $C=74$, we have
$\begin{array}{ll}
74=k_1+6k_2 & \ldots \unicode{x2461}
\end{array}$$\unicode{x2461} – \unicode{x2460}$, we have
$\begin{array}{rcl}
12 & = & 4 k_2 \\
k_2 & = & 3
\end{array}$Sub. $k_2=3$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
62 & = & k_1 + 2(3) \\
k_1 & = & 56
\end{array}$Therefore, $C=56+3A$. For $A=13$, we have
$\begin{array}{rcl}
C & = & 56+3(13) \\
& = & 95
\end{array}$Therefore, the required cost is $\$95$.
- Let $V_s\mbox{ m}^3$ be the volume of the smaller can. Let $A_l\mbox{ m}^2$ be the surface area of the larger can. Then, the volume of the larger can $=8V_s\mbox{ m}^3$.
Since the two cans are similar, we have
$\begin{array}{rcl}
\dfrac{A_l}{13} & = & \left(\sqrt[3]{\dfrac{8V_s}{V_s}}\right)^2 \\
A_l & = & 52
\end{array}$Therefore, the required cost
$\begin{array}{cl}
= & 56+3(52) \\
= & \$212
\end{array}$
2012-I-11
Ans: (a) $\$95$ (b) $\$212$
