Question

Find the critical points and test for relative extrema. List the critical points for which the Second-Partials Test fails. $$ f(x, y)=\left(x^{2}+y^{2}\right)^{2 / 3} $$

Short answer

The only critical point of the function is at (0,0), but it is not defined by the Second-Partials Test due to the undefined nature of the second derivatives at this point.

Step-by-step solution

Find the Partial Derivatives

The first partial derivatives of the function with respect to x and y are calculated as follows: \[f_x=\frac{4}{3}x\left(x^{2}+y^{2}\right)^{-1 / 3}\] and \[f_y=\frac{4}{3}y\left(x^{2}+y^{2}\right)^{-1 / 3}\].

Find the Critical Points

The critical points occur where both partial derivatives are equal to 0 or are undefined. Setting both \(f_x\) and \(f_y\) to 0: \[\frac{4}{3}x\left(x^{2}+y^{2}\right)^{-1 / 3}=0\] and \[\frac{4}{3}y\left(x^{2}+y^{2}\right)^{-1/3}=0\]. The only solution is at (x, y) = (0, 0), which is a critical point.

Apply the Second-Partials Test

The second partials test requires the computation of the second derivatives of the function, including \(f_{xx}, f_{yy},\) and \(f_{xy}\). These are: \[f_{xx}=\frac{4}{3}(1-\frac{x^{2}}{x^{2}+y^{2}})\left(x^{2}+y^{2}\right)^{-1 / 3}, f_{yy}=\frac{4}{3}(1-\frac{y^{2}}{x^{2}+y^{2}})\left(x^{2}+y^{2}\right)^{-1 / 3},\] and \(f_{xy}=\frac{4}{3}(-\frac{xy}{x^{2}+y^{2}})\left(x^{2}+y^{2}\right)^{-1 / 3}\). The discriminant D at a point (x, y) is defined as \(D = f_{xx}f_{yy} - (f_{xy})^{2}\). At the critical point (0,0), these are undefined, and hence the Second-Partials Test fails. Therefore, the critical point (0,0) is undefined for the Second-Partials Test.