If m is the minimum value of k for which the function $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}\sqrt{\mathrm{kx}-{\mathrm{x}}^2}$ is increasing in the interval $\left[0, 3\right]$ and $\mathrm{M}$ is the maximum value of $\text{⨍ }$ in $\left[0, 3\right]$ when $\mathrm{k}=\mathrm{m},$ then the ordered pair $\left(\mathrm{m}, \mathrm{M}\right)$ is equal to:

$$\begin{gathered} f\left(x\right)=x\sqrt{kx-x^2} \\ f\text{'}\left(x\right)=\sqrt{kx-x^2}+x\frac{1}{2\sqrt{kx-x^2}}(k-2x) \\ =\frac{2kx-2x^2+kx-2x^2}{2\sqrt{kx-x^2}} \\ =\frac{3kx-4x^2}{2\sqrt{kx-x^2}} \\ =\frac{-x(4x-3k)}{2\sqrt{kx-x^2}} \\ f\left(x\right) is increa\sin g \\ f\text{'}\left(x\right)>0 \\ -x(4x-3k)>0 \\ x(4x-3k)<0 \\ x\in \left(0,\frac{3k}{4}\right) \\ given x\in (0,3) \\ \Rightarrow k=4 \\ m=k=4 \\ f\left(x\right) is increa\sin g \left[0,3\right] \\ m=maximum value=f\left(3\right)=3\sqrt{12-9}=3\sqrt{3} \end{gathered}$$