Relative Minimums and Maximums — Question 8

Question 8 – Solution

Strategy. Restrict the function to coordinate lines through the interior point and apply the one-variable necessary condition.

Step 1: Necessary condition Suppose $f$ has a relative extremum at $(a,b)$. The one-variable restriction $$\phi(x)=f(x,b)$$ has a relative extremum at $x=a$. Differentiability gives $$f_x(a,b)=\phi'(a)=0.$$ Likewise, $\psi(y)=f(a,y)$ has a relative extremum at $b$, so $$f_y(a,b)=\psi'(b)=0.$$ Therefore $$\boxed{\nabla f(P)=\mathbf 0}.$$

Step 2: The converse A zero gradient makes $P$ a critical point, not necessarily an extremum. For instance, $f(x,y)=x^2-y^2$ has gradient zero at the origin but takes both positive and negative values nearby, so the origin is a saddle.

Step 3: Application For $g(x,y)=x+e^y$, $$g_x=1,\qquad g_y=e^y.$$ In particular, the gradient is never zero. Since every point of $\mathbb R^2$ is interior and $g$ is differentiable everywhere, the necessary condition rules out a relative maximum or minimum at every point. Hence $$\boxed{g\text{ has no relative extrema on }\mathbb R^2}.$$