$\def\D{\displaystyle}$(CIE 0606/2012/s/11/Q11 and Solution) The diagram shows part of the curve $\D y = 9x^2 - x^3,$ which meets the x-axis at the origin $\D O$ and at the point $\D A$. The line $\D y - 2x + 18 = 0$ passes through $\D A$ and meets the y-axis at the point $\D B.$
(i) Show that, for $\D x \ge 0, 9x^2 - x^3 \le 108.$ [4] (ii) Find the area of the shaded region bounded by the curve, the line $\D AB$ and the y-axis. [6]
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*********math solution*************
$y=9 x^{2}-x^{3}$
$\dfrac{d y}{d x}=18 x-3 x^{2}$
For stationary points, $\dfrac{d x}{d x}=0$. Thus $18 x-3 x^{2}=0$
$x(18-3 x)=0$
$x=0$ or $x=6$
$\dfrac{d^{2} y}{d x^{2}}=18-6 x$
When $x=0, \dfrac{d^{2} y}{d x^{2}}=18>0$
and $\quad y=0$
$\therefore(0,0)$ is minimum point
At $x =6, \dfrac{d^{2} y}{d x^{2}}=18-6 \times 6=-18<0 .$
and $y =9(6)^{2}-(6)^{3}=108 .$
$\therefore(6,108)$ is maximum point.
Thus. $9 x^{2}-x^{3} \leqslant 108.$
ii) When $y=0, \quad 0-2 x+18=0$
$\therefore x=9$
$\therefore A=(9,0) .$
Area of shaded region $=\displaystyle\int_{0}^{9}\left(9 x^{2}-x^{3}\right)-(2 x-18) d$
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