91Ó°ÊÓ

Calculate the first partial derivatives of both functions with respect to x and y to form the gradient. For \(f(x,y)=-\left(x^{2}-1\right)^{2}-\left(x^{2}-e^{y}\right)^{2}\): \[\nabla f(x,y) = \begin{bmatrix}\frac{\partial{f}}{\partial{x}} \\ \frac{\partial{f}}{\partial{y}}\end{bmatrix} = \begin{bmatrix}-4x\left(x^{2}-1\right)-4x\left(x^{2}-e^{y}\right) \\ 2\left(x^{2}-e^{y}\right)e^{y}\end{bmatrix}\] For \(f(x,y)=4x^{2}e^{y}-2x^{4}-e^{4y}\): \[\nabla f(x,y) = \begin{bmatrix}\frac{\partial{f}}{\partial{x}} \\ \frac{\partial{f}}{\partial{y}}\end{bmatrix} = \begin{bmatrix}8xe^{y}-8x^{3} \\ 4x^{2}e^{y}-4e^{4y}\end{bmatrix}\]

Calculate the second partial derivatives of both functions with respect to x and y to form the Hessian matrix. For \(f(x,y)=-\left(x^{2}-1\right)^{2}-\left(x^{2}-e^{y}\right)^{2}\): \[\text{Hessian}(f) = \begin{bmatrix}\frac{\partial^2{f}}{\partial{x}^2} & \frac{\partial^2{f}}{\partial{x}\partial{y}} \\ \frac{\partial^2{f}}{\partial{y}\partial{x}} & \frac{\partial^2{f}}{\partial{y}^2}\end{bmatrix} = \begin{bmatrix}-12x^2+4+4e^{y} & 8xe^{y} \\ 8xe^{y} & -2e^{y}\left(x^{2}-e^{y}\right)\end{bmatrix}\] For \(f(x,y)=4x^{2}e^{y}-2x^{4}-e^{4y}\): \[\text{Hessian}(f) = \begin{bmatrix}\frac{\partial^2{f}}{\partial{x}^2} & \frac{\partial^2{f}}{\partial{x}\partial{y}} \\ \frac{\partial^2{f}}{\partial{y}\partial{x}} & \frac{\partial^2{f}}{\partial{y}^2}\end{bmatrix} = \begin{bmatrix}8e^{y}-24x^2 & 8x \\ 8x & 8x^{2}e^{y}-16e^{4y}\end{bmatrix}\]

Solve the gradient vector equation for each function, setting the gradient equal to the zero vector to find the critical points. For \(f(x,y):-4x\left(x^{2}-1\right)-4x\left(x^{2}-e^{y}\right)=0\), and \( 2\left(x^{2}-e^{y}\right)e^{y}=0\). The solutions to these simultaneous equations are given by the critical points \((\pm 1, 0)\). For \(f(x,y): 8xe^{y}-8x^{3}=0\), and \(4x^{2}e^{y}-4e^{4y}=0\). The solutions to these simultaneous equations are given by the critical points \((\pm \frac{\sqrt{2}}{2}, 0)\).

Evaluate the Hessian of each function at its critical points, and determine the nature of the critical points by studying the signs of the eigenvalues. For the critical points \((\pm 1, 0)\), \(\text{Hessian}(f) = \begin{bmatrix}-8 & 0 \\ 0 & 0\end{bmatrix}\). The eigenvalues of this matrix are \(-8\) and \(0\). Since one eigenvalue is negative and the other is zero, we can't determine the nature of these critical points using the Hessian test. However, because both eigenvalues are non-positive, we know that these points are either local maxima or saddle points. Since we are given that there are no saddle points in the problem statement, we can conclude that \((\pm 1, 0)\) are local maxima for \(f(x, y)\). For the critical points \((\pm \frac{\sqrt{2}}{2}, 0)\), \(\text{Hessian}(f) = \begin{bmatrix}4 & \pm 2\sqrt{2} \\ \pm 2\sqrt{2} & 2\end{bmatrix}\). The eigenvalues of this matrix are \(3 \pm \sqrt{2}\). Both eigenvalues are positive, implying that \((\pm \frac{\sqrt{2}}{2}, 0)\) are local maxima for \(f(x, y)\). In conclusion, the two functions have two local maxima but no other extreme points.