If $\mathrm{A}=\left\{\mathrm{x}/9\mathrm{x}\ge {\mathrm{x}}^2+20\right\} \mathrm{and} \mathrm{f}:\mathrm{A}\to \mathrm{R} \mathrm{is}$defined by $\mathrm{f}\left(\mathrm{x}\right)=2{\mathrm{x}}^3-15{\mathrm{x}}^2+36\mathrm{x}-48$ then the maximum value of f(x) is 

$\begin{array}{r}{x}^2+20\le 9x\\ x^2-9x+20\le 0\\ x^2-4x-5x+20\le 0\\ x(x-4)-5(x-4)\le 0\\ (x-4)(x-5)\le 0\\ 4\le x\le 5\end{array}$

$\mathrm{f}\left(\mathrm{x}\right)=2{\mathrm{x}}^3-15{\mathrm{x}}^2+36\mathrm{x}-48$

Diff w.r.t.x

$\begin{array}{r}{f}^{\mathrm{\prime }}\left(x\right)=6x^2-30x+36=6\left(x^2-5x+6\right)\\ =6(x-2)(x-3)\end{array}$

f(x) is increasing in $(-\mathrm{\infty },2)\cup (3,\mathrm{\infty })$

Maximum value of f(x) at x=5

$\begin{array}{r}f\left(5\right)=2(5{)}^3-15(5{)}^2+36\left(5\right)-48\\ =250-375+180-48\\ =430-423=7\end{array}$