Question

Examine the function for relative extrema and saddle points. $$ g(x, y)=x y $$

Short answer

(0,0) is a saddle point of the function \(g(x, y)=x y\)

Step-by-step solution

Compute the First Partial Derivatives

For any function of two variables, start by finding its first-order partial derivatives with respect to x and y:The first partial derivative of \(g\) with respect to \(x\) is \(g_x(x, y) = y\). The first partial derivative of \(g\) with respect to \(y\) is \(g_y(x, y) = x.\)

Set Both First Partial Derivatives to Zero

Setting \(g_x = y = 0\) and \(g_y = x = 0\), any point that makes both of these true is a critical point. In this case, it's evident that \((0,0)\) is the only critical point.

Compute the Second Partial Derivatives

The second partial derivatives are:\(g_{xx}(x, y) = 0\), \(g_{yy}(x, y) = 0\), and \(g_{xy}=g_{yx} = 1\)

Apply the Second Partials Test

The second partials test uses the discriminant \(D = g_{xx}g_{yy} - g_{xy}^2\). If \(D > 0\) and \(g_{xx}\) (or \(g_{yy}\)) is positive at the critical point, the function has a local minimum. If \(D > 0\) and \(g_{xx}\) (or \(g_{yy}\)) is negative, the function has a local maximum. If \(D < 0\), the function has a saddle point. For the test to be inconclusive, \(D = 0\).For the critical point (0,0), \(D = g_{xx}(0,0)g_{yy}(0,0) - g_{xy}(0,0)^2 = 0(0) - 1^2 = -1<0\)

Determine the nature of the Critical Point

Because the discriminant \(D\) is less than zero at the critical point \((0,0)\), this means that \((0,0)\) is a saddle point of the function \(g(x, y)=x y\).