Question

Use a computer algebra system to graph the surface and locate any relative extrema and saddle points. $$ z=e^{x y} $$

Short answer

The graph of \(z=e^{x y}\) has one critical point at (0,0), which is a saddle point. There are no relative extrema.

Step-by-step solution

Graphing the Surface

Initially, the surface should be graphed using a computer algebra system. The three-dimensional function \(z=e^{x y}\) demonstrates how the values of z changes with different x and y values. This is beneficial to get a general idea of where the high and low points on the graph are.

Calculating the partial derivatives

For finding relative extrema or saddle points, first, determine the critical points. To find the critical points, start to take the partial derivatives and set them equal to zero. The first partial derivatives are: \( \frac{\partial z}{\partial x} = y e^{xy} \) and \( \frac{\partial z}{\partial y} = x e^{xy} \). Next, set them equal to zero and solve for x and y.

Identifying the critical points

Upon setting the first partial derivatives to zero, we get y=0 from \(\frac{\partial z}{\partial x}\) equation and x=0 from \(\frac{\partial z}{\partial y}\) equation. Hence, the only critical point is at (0,0).

Calculating the second partial derivatives

After finding the critical points, it's time to classify them as relative extrema or saddle points. For this, compute the second partial derivatives: \(\frac{\partial^2 z}{\partial x^2} = y^2 e^{xy} \), \(\frac{\partial^2 z}{\partial y^2} = x^2 e^{xy} \) and \(\frac{\partial^2 z}{\partial x \partial y} = e^{xy} + xy e^{xy} \).

Applying the second derivative test

Apply the second derivative test by finding the determinant of the Hessian matrix. The Hessian matrix is: H = \( \begin{bmatrix} \frac{\partial^2 z}{\partial x^2} && \frac{\partial^2 z}{\partial x \partial y} \\ \frac{\partial^2 z}{\partial x \partial y} && \frac{\partial^2 z}{\partial y^2} \end{bmatrix} = \begin{bmatrix} y^2 e^{xy} && e^{xy} + xy e^{xy} \\ e^{xy} + xy e^{xy} && x^2 e^{xy} \end{bmatrix} \) at (0,0). The determinant is \(\Delta = y^4 e^{2xy} - y^2 e^{2xy} - x^2 e^{2xy} + 2x^2 y^2 e^{2xy} \). The determinant must be positive for a relative extrema and negative for a saddle point.

Representing the results

Substitute the critical point into the determinant. For (0,0), determinant \(\Delta = 0-0 = 0\). As it must be positive for a relative extrema and negative for a saddle point, \(\Delta = 0\) indicates that the test is inconclusive. Therefore, by inspecting the graph, (0,0) is not a maximum or minimum point but appears to be a saddle point.