2013-I-19

1. The total floor area of all public housing flats at the end of the 2nd year

   $\begin{array}{cl}
   = & 9\times10^6 (1+ r\%) – 3\times 10^5 \text{ m}^2
   \end{array}$

2. The total floor area of all public housing flats at the end of the 3rd year

   $\begin{array}{cl}
   = & (9\times10^6 (1+ r\%) – 3\times 10^5) \times(1+r\%)-3\times10^5 \\
   = & 9\times10^6 (1+r\%)^2 -3\times10^5[1+(1+r\%)] \text{ m}^2
   \end{array}$

   Since the total floor area of all public housing flats at the end of the 3rd year is $1.026\times 10^7\text{ m}^2$, then we have

   $\begin{array}{rcl}
   9\times10^6 (1+r\%)^2 -3\times10^5[1+(1+r\%)] & = & 1.026\times10^7 \\
   9\times10^6(1+r\%)^2 -3\times10^5(1+r\%) -1.056\times10^7 & = & 0
   \end{array}$

   $\therefore 1+r\% = 1.1 \text{ or } 1+r\% = -1.07 \text{(rejected)} $

   Therefore, $r=10$.

1. The total floor area of all public housing flats at the end of the $n$th year

   $\begin{array}{cl}
   = & [9\times 10^6(1.1)^{n-1} -3\times 10^5(1.1)^{n-2} -\ldots – 3\times10^5(1.1) -3\times10^5 ] \\
   = & 9\times10^6(1.1)^{n-1} -\dfrac{3\times10^5(1-(1.1)^{n-1}}{1-1.1} \\
   = & 6\times10^6(1.1)^{n-1} +3\times 10^6 \text{ m}^2
   \end{array}$

2. By the result of (b)(i), we have

   $\begin{array}{rcl}
   6\times10^6(1.1)^{n-1} +3\times 10^6 & > & 4 \times 10^7 \\
   6(1.1)^{n-1} & > & 37 \\
   (1.1)^{n-1} & > & \dfrac{37}{6} \\
   \log (1.1)^{n-1} & > & \log \dfrac{37}{6} \\
   n-1 & > & \dfrac{\log\frac{37}{6}}{\log 1.1} \\
   n & > & 20.086~717~15
   \end{array}$

   Therefore, at the end of the 21st year, the total floor area of all public housing flats first exceed $4\times10^7\text{ m}^2$.

$\begin{array}{rcl}
a(1.21)^1 + b & = & 1\times10^7 \\
1.21a +b & = & 1\times 10^7\ \ldots \unicode{x2460}
\end{array}$

For $n=2$,

$\begin{array}{rcl}
a(1.21)^2 + b & = & 1.063\times10^7 \\
1.4641a+b & = & 1.063\times10^7 \ \ldots \unicode{x2461}
\end{array}$

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

$\begin{array}{rcl}
0.2541 a & = & 6.3\times 10^5 \\
a & = & 2.479~338~843 \times 10^6
\end{array}$

Sub. $a=2.479~338~843$ into $\unicode{x2460}$, we have

$\begin{array}{rcl}
1.21(2.479~338~843\times10^6) + b & = & 1\times 10^7 \\
b & = & 7\times10^6
\end{array}$

Therefore, the total floor area of public housing flats needed at the end of the $n$th year is $(2.479~338~843\times10^6(1.21)^n+7\times10^6)\text{ m}^2$.

Let $d(n)$ be the difference between the total floor built and the total floor needed at the end of the $n$th year.

$\begin{array}{rcl}
d(n) & = & (6\times10^6(1.1)^{n-1} +3\times10^6) -(2.479~338~843\times10^6(1.21)^n+7\times10^6) \\
& = & -3\times10^6(1.1)^{2(n-1)} +6\times10^6(1.1)^{n-1} -4\times10^6 \\
\end{array}$

Consider the discriminant of the equation $d(n)=0$, we have

$\begin{array}{rcl}
\Delta & = & (6\times10^6)^2 – 4 (-3\times10^6)(-4\times10^6) \\
& = & -1.2 \times 10^{13} \\
& < & 0
\end{array}$

Therefore, there is no real roots for the equation $d(n)=0$. Note that the coefficient of $(1.1)^{2(n-1)}$ is negative and the discriminant is also negative. Therefore, $d(n)<0$.

Hence, the total floor built is always smaller than the total floor needed. I don’t agree.