jmc

We write $$x\sqrt{12-x}+\sqrt{12x-x^3}=\sqrt{12-x}\cdot \sqrt{x^2}+\sqrt{x}\cdot \sqrt{12-x^2}$$By Cauchy-Schwarz, $$(\sqrt{12-x}\cdot \sqrt{x^2}+\sqrt{x}\cdot \sqrt{12-x^2}{)}^2\le (12-x+x)(x^2+12-x^2)=144,$$so $$\sqrt{12-x}\cdot \sqrt{x^2}+\sqrt{x}\cdot \sqrt{12-x^2}\le 12.$$But $\sqrt{12-x}\cdot \sqrt{x^2}+\sqrt{x}\cdot \sqrt{12-x^2}\ge 12,$ so the expression must be equal to 12. From the equality condition for Cauchy-Schwarz, $$\frac{12-x}{x}=\frac{x^2}{12-x^2}.$$Then $(12-x)(12-x^2)=x^3,$ which simplifies to $x^2+x-12=0.$ This factors as $(x-3)(x+4)=0,$ so the only solution is $x=\boxed{3}.$