For a pair of straight lines represented by $$ax^2+by^2+2gx+2fy+2hxy+c=0,$$ point of intersection can be determined by solving equations: $\frac{\partial z}{\partial x}=0$ and $\frac{\partial z}{\partial y}=0$.
[where $z=ax^2+by^2+2gx+2fy+2hxy+c$]
But to find the minima/maxima for the multivariable function $z=f(x,y)$, we have to solve the exact same two equations i.e. $\frac{\partial z}{\partial x}=0$ and $\frac{\partial z}{\partial y}=0$.
And if by solving the two equations we get some value of $(x,y)$ say $(x_0,y_0)$. Then, $f(x_0,y_0)$ should give us minimum/maximum value of $z$, but it turns out it always gives $z=0$.
How to find the extremum value of $z$? Where am I wrong?
Finding the extreme value of a multivariable function.
analytic geometrymaxima-minimamultivariable-calculus
Best Answer
Your surface $z=f(x,y)$ is a hyperbolic paraboloid through two lines on the $XY$ plane. By setting the partial derivatives equal to zero you get the saddle point of the surface that is clearly the point of intersection of your lines and hence on the $XY$ plane where $z=0$. There is no extremum.