2012-I-19

$\left\{ \begin{array}{ll}
254~100 = ab^2 & \ldots \unicode{x2460} \\
307~461 = ab^4 & \ldots \unicode{x2461}
\end{array} \right.$

$\unicode{x2461} \div \unicode{x2460}$, we have

$\begin{array}{rcl}
1.21 & = & b^2 \\
b & = & 1.1
\end{array}$

Sub. $b=1.1$ into $\unicode{x2460}$, we have

$\begin{array}{rcl}
254~100 & = & a (1.1)^2 \\
a & = & 210~000
\end{array}$

Hence, the required weight of the goods

$\begin{array}{cl}
= & 210~000(1.1)^{2(4)} \\
= & 450~153.6501 \mbox{ tonnes}
\end{array}$

$\begin{array}{cl}
= & ab^{2} + ab^{4} + \cdots + ab^{2n} \\
= & \dfrac{ab^2((b^2)^n-1)}{b^2-1} \\
= & \dfrac{210~000(1.1)^2((1.1)^{2n}-1)}{1.1^2-1} \\
= & 1~210~000[(1.1)^{2n}-1] \mbox{ tonnes}
\end{array}$

$\begin{array}{rcl}
B(m) & = & 2(210~000)(1.1)^m
\end{array}$

and

$\begin{array}{rcl}
A(m+4) & = & (210~000)(1.1)^{2m+8}
\end{array}$

Hence, we have

$\begin{array}{rcl}
\dfrac{A(m+4)}{B(m)} & = & \dfrac{210~000(1.1)^{2m+8}}{2(210~000)(1.1)^m} \\
& = & \dfrac{(1.1)^{m+8}}{2} \\
& = & (1.1)^m \times \dfrac{(1.1)^8}{2} \\
& > & (1.1)^m \\
& > & 1 \\
A(m+4) & > & B(m)
\end{array}$

Therefore, I agree the claim.

$\begin{array}{cl}
= & 2ab +2ab^2 + \cdots + 2ab^m \\
= & \dfrac{2ab(b^m-1)}{b-1} \\
= & \dfrac{2(210~000)(1.1)[(1.1)^m-1]}{1.1-1} \\
= & 4~620~000[(1.1)^m-1] \mbox{ tonnes}
\end{array}$

Therefore, we have

$\begin{array}{rcl}
1.21\times10^6[(1.1)^{2m+8}-1]+ 4.62\times10^6[(1.1)^m-1] & > & 2\times 10^7 \\
(1.1)^{10} (1.1)^{2m} + 4.62(1.1)^m -25.83 & > & 0 \\
\end{array}$

Therefore, we have

$1.1^m>2.388~382~716$ or $1.1^m<-4.169~592~713$ (rejected).

Hence, we have

$\begin{array}{rcl}
1.1^m & > & 2.388~382~716 \\
m\log 1.1 & > & \log 2.388~382~716 \\
m & > & 9.134~558~877
\end{array}$

The least value of $m$ is $10$. Therefore, the new facilities should be installed in the $10$th year since the start of $Y$. i.e. The new facilities should be installed in the $14$th year since the start of $X$.