7. Find the maximum value of the function $f(x)=\sqrt{5 x}+\sqrt{6-x}$.


7. Let $a=(\sqrt{5}, 1), b=(\sqrt{x}, \sqrt{6-x})$, then $f(x)=a \cdot b,|a| \cdot|b|=6$, and $a \cdot b \leqslant|a| \cdot|b|$, so $f(x) \leqslant 6$. Equality holds if and only if $b=\lambda a(\lambda>0)$, hence from $\frac{\sqrt{x}}{\sqrt{5}}=\frac{\sqrt{6-x}}{1}=\lambda>0$, we get $x=5, \lambda=1$. Therefore, when $x=5$, $f(x)_{\max }=6$.