# Example

Consider the function $z(x,y)=3x^2+2xy+y^3$ where $x\geq 0$ and $y\geq 1$.

The first-order partial derivatives are given by
1. $z'_x(x,y)=6x+2y$;
2. $z'_y(x,y)=2x+3y^2$.
The second-order partial derivatives are given by
1. $z''_{xx}(x,y)=6$;
2. $z''_{yy}(x,y)=6y$;
3. $z''_{xy}(x,y)=z''_{yx}(x,y)=2$.
From the above it follows that the criterion function is given by $C(x,y)=6\cdot 6y-(2)^2=36y-4$.

Since $C(x,y)\geq 0$, $z''_{xx}(x,y)\geq 0$ and $z''_{yy}(x,y)\geq 0$, for $x\geq 0$ and $y\geq 1$, it follows from the second-order condition that the function $z(x,y)$ is convex on the part of the domain where $x\geq 0$ and $y\geq 1$.