I´m having a lot of trouble showing if the objective function is quasiconvex or not. I already tried with the bordered Hessian but it gets really messy :-/. Maybe someone can help me with this! Thanks in advance!
The problem: $$max f(x,y)=x^{1/2}*y^{1/3}$$
subject to:$$3x+4y ≤25$$
Make up the generalized Lagrange function:
$L=x^{\frac{1}{2}}y^{\frac{1}{3}}+\lambda(25-3x-4y).$
Kuhn-Tucker conditions for Maximum:
$L_x=\frac{1}{2}x^{-\frac{1}{2}}y^{\frac{1}{3}}-3\lambda\leq0,\quad x\geq0, \quad xL_x=0;$
$L_y=\frac{1}{3}x^{\frac{1}{2}}y^{-\frac{2}{3}}-4\lambda\leq0,\quad y\geq0, \quad yL_y=0;$
$L_{\lambda}=25-3x-4y\geq0,\quad \lambda\geq0, \quad \lambda L_{\lambda}=0.$
Cases:
1) $x,y,\lambda>0 \Rightarrow \begin{cases} L_x=0, \\ L_y=0, \\ L_{\lambda}=0 \end{cases} \Rightarrow (x,y)=(5,2.5)\Rightarrow f=5^{\frac{5}{6}}\cdot2^{-\frac{1}{3}} (max).$
2) $x=0,y>0\Rightarrow f=0.$
3) $x>0,y=0\Rightarrow f=0.$
4) $x=0,y=0\Rightarrow f=0.$