Problem 25 - College Algebra and Calculus: An Applied Approach

Question

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

Step-by-Step Solution

Answer

The critical point of the function \(f(x, y)=(x y)^{2}\) is (0,0). The Second-Partials Test fails at this point.

Step 1: Calculate the Partial Derivatives

The critical points are found by setting the first partial derivatives equal to zero. Start by calculating the partial derivative of the function with respect to x, denoted as \(f_x\), and with respect to y, denoted as \(f_y\). The given function is \(f(x, y)=(x y)^{2}\). Hence, \(f_x=2xy^2\) and \(f_y=2x^2y\).

Step 2: Find the Critical Points

After computing the partial derivatives, we need to find the points that will make both \(f_x\) and \(f_y\) equal to zero. That implies solving the equations \(2xy^2=0\) and \(2x^2y=0\). Solving these gives the singular solution, that is (x,y)=(0,0).

Step 3: Apply the Second-Partials Test

In the next step, apply the Second-Partials Test using the second order partial derivatives denoted as \(f_{xx}\), \(f_{yy}\), and \(f_{xy}\). These are calculated as: \(f_{xx}=2y^2\), \(f_{yy}=2x^2\), and \(f_{xy}=4xy\). The Second-Partials Test uses the formula \(D=f_{xx}f_{yy} - (f_{xy})^2\) where D is used to determine the type of critical point. Calculate D at (0,0) gives D = 0.

Step 4: Determine the Extrema or Failures

Substitute the critical points into the formula from the second partials test. If D>0 and \(f_{xx}<0\) or \(f_{yy}<0\), then the critical point is a local maximum. If D >0 and \(f_{xx}>0\) or \(f_{yy}>0\), then the critical point is a local minimum. If D<0, then it's a saddle point. If D=0, the test is inconclusive. Here, D is 0 at the point (0,0), therefore the Second-Partials Test fails at this point.