The maximum value of $\mathrm{f}\left(\mathrm{x}\right)=2{\mathrm{x}}^3-21{\mathrm{x}}^2+36\mathrm{x}+20$  in the interval $0\le \mathrm{x}\le 2$  is _____

$\begin{array}{l}\mathrm{Given} \mathrm{f}\left(\mathrm{x}\right)=2x^3-21x^2+36x+20\\ diff w.r.to x on both sides\\ f\text{'}\left(x\right)=6{\mathrm{x}}^2-42\mathrm{x}+36\\ for max or min f\text{'}\left(x\right)=0\\ \Rightarrow 6{\mathrm{x}}^2-42\mathrm{x}+36=0\\ \Rightarrow {\mathrm{x}}^2-7\mathrm{x}+6=0\\ \Rightarrow \mathrm{x}=1,\mathrm{x}=6\\ Now f\left(1\right)=37\\ f\left(0\right)=20\\ f\left(2\right)=24\\ \therefore maximum value of f\left(x\right)=37(\because 0\le x\le 2 so we can\text{'}t take x value is 6)\\ \end{array}$