2010-I-10 – Solving Master

Ans: (a)

C = 4 x + 5 x^{2}

(b)

\frac{36}{5} m

Let $C = k_{1} x + k_{2} x^{2}$ , where $k_{1}$ and $k_{2}$ are non-zero constants. For $x = 4$ and $C = 96$ , we have
$\begin{array}{rcl} 96 & = & k_{1} (4) + k_{2} (4)^{2} \\ 24 & = & k_{1} + 4 k_{2} \dots ① \end{array}$

For $x = 5$ and $C = 145$ , we have

$\begin{array}{rcl} 145 & = & k_{1} (5) + k_{2} (5)^{2} \\ 29 & = & k_{1} + 5 k_{2} \dots ② \end{array}$

$② - ①$ , we have

$\begin{array}{rcl} 5 & = & k_{2} \\ k_{2} & = & 5 \end{array}$

Sub. $k_{2} = 5$ into $①$ , we have

$\begin{array}{rcl} 24 & = & k_{1} + 4 (5) \\ k_{1} & = & 4 \end{array}$

Therefore, $C = 4 x + 5 x^{2}$ .
For $C = 288$ , we have
$\begin{array}{rcl} 288 & = & 4 x + 5 x^{2} \\ 5 x^{2} + 4 x - 288 & = & 0 \\ (5 x - 36) (x + 8) & = & 0 \\ ∴ x = \frac{36}{8} & or & x = - 8 (rejected) \end{array}$

Therefore, the perimeter is $\frac{36}{8} m$ .