A production function for a firm is defined by
$ P(x ,y ,z) = x^{\frac{1}{2}}ln y - z^{2} $
Find the set of values of y for which P(x,y,z) is a strictly concave function
$\Delta p = ( \frac{1}{2}x^{\frac{-1}{2}} lny , x^{\frac{1}{2}} \frac{1}{y} , -2z)$ I know that a function is strictly concave when its Hessian matrix is negative definite.
I have computed the Hessian matrix $$H (x,y,z)= \begin{pmatrix} \frac{-1}{4}x^{\frac{-3}{2}} lny & \frac{1}{2} x^{\frac{-1}{2}} \frac{1}{y} & 0 \\ \frac{1}{2} x^{-\frac{1}{2}} \frac{1}{y} & -x^{-\frac{1}{2}} & 0 \\ 0 & 0 & -2 \end{pmatrix} $$
A negative definite Hessian is a sufficient (but not necessary) condition for strict concavity. But here this will do. I assume $x,y,z \ge 0$ because we are dealing with a production function.
The Hessian is negative definite if its leading principal minors alternate in sign. The correct Hessian for your function (yours had a typo in $h_{22}$) is $$H (x,y,z)= \begin{pmatrix} -\dfrac{x^{-3/2}\ln y}{4} & \dfrac{x^{-1/2}}{2y} & 0 \\ \dfrac{x^{-1/2}}{2y} & - \dfrac{x^{1/2}}{y^2} & 0 \\ 0 & 0 & -2 \end{pmatrix} $$
The first leading principal minor $$-\frac{x^{-3/2}\ln y}{4} < 0$$ implies $x>0$ and $y>1$. The second leading principal minor $$\dfrac{x^{-1}}{4y^2} \left(\ln y - 1\right) >0$$ implies $y > e$. Finally, the third leading principal minor being strictly negative adds no further restrictions.
Combining the above, we get $x>0$ and $y>e$.