Ans: (a) $(18,324)$ (b) (i) $A=\dfrac{3}{2}(36x-x^2)\text{ m}^2$ (ii) No
- By the method of completing the square, we have$\begin{array}{rcl}
f(x) & = & 36x-x^2 \\
& = & -(x^2-36x + (\dfrac{-36}{2})^2 – (\dfrac{-36}{2})^2) \\
& = & -(x-18)^2 +324
\end{array}$Therefore, the vertex of the graph of $y=f(x)$ is $(18,324)$.
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- The length of the rectangular restricted zone$\begin{array}{cl}
= & \dfrac{1}{2} (108-3x) \text{ m}
\end{array}$Hence, we have
$\begin{array}{rcl}
A & = & \dfrac{1}{2}(108-3x) \times x \\
& = & \dfrac{3}{2}(36x-x^2) \text{ m}^2
\end{array}$ - By the result of (a), the maximum value of $f(x)=36x-x^2$ is $324$.Therefore, the maximum value of $A$
$\begin{array}{cl}
= & \dfrac{3}{2} \times 324 \\
= & 486 \text{ m}^2 \\
< & 500 \text{ m}^2
\end{array}$Therefore, the area of this restricted zone cannot be greater than $500\text{ m}^2$. I don’t agree.
- The length of the rectangular restricted zone$\begin{array}{cl}