Finding Critical Points

Consider the hyperbolic paraboloid defined by the equation \( z=f(x,y)=y^{2}-x^{2}\). Show that \((0,0)\) is the only critical point of \(f\), but that \(f\) has neither a local maximum nor a local minimum at \((0,0)\).

Solution  The partial derivatives of \(f\) are \[ f_{x}(x,y)=-2x\qquad \hbox{and} \qquad f_{y}(x,y)=2y \]

At \(( 0,0) \), both partial derivatives equal \(0\), so \((0,0)\) is the only critical point.

See Figure 15. If we consider the values of \(f(x,0)=-x^{2}\), the function attains a maximum value of \(0\) at the origin. However, if we consider the values of \(f(0,y)=y^{2}\), the function attains a minimum value at the origin. In other words, at the critical point \((0,0)\), the function appears to have a maximum when viewed in one direction and to have a minimum when viewed in another direction. As a result, \(z=f(x,y)=y^{2}-x^{2}\) has neither a maximum nor a minimum at (\(0,0\)).