Consider the following function with domain #x \gt 0\land y \gt 0#: \[f(x,y)=3\cdot y+x^{a}\cdot \sqrt{y}-2\cdot x+3\tiny,\] where #a# is a constant.

The interval of values of #a# for which the conditions of the second order derivative test for #f# to be concave are satisfied, is of the form #0\le a \le c#. Determine the constant #c#.

You can use the following expressions for the second derivatives and the Hessian of #f#:
\[\begin{array}{rcl} f_{xx} &=& \left(a^2-a\right)\cdot x^{a-2}\cdot \sqrt{y} \\
f_{xy}&=& {{a\cdot x^{a-1}}\over{2\cdot \sqrt{y}}} \\
f_{yy}&=& -{{x^{a}}\over{4\cdot y^{{{3}\over{2}}}}} \\
f_{xx} \cdot f_{yy} - f_{xy}^2 &=& -{{a\cdot \left(2\cdot a-1\right)\cdot x^{2\cdot a-2}}\over{4\cdot y}}
\end{array} \]