Final answer:
The critical point of the function f(x, y) = 2 - 7x² - 8y² is at (0, 0), and the second derivative test classifies this point as a local maximum since the determinant from the test is positive and the second partial derivatives at the point are negative.
Step-by-step explanation:
To find the critical point of the function f(x, y) = 2 - 7x² - 8y², we need to find the points where the partial derivatives with respect to both x and y are zero.
The partial derivative with respect to x is f_x(x, y) = -14x, and setting this equal to zero gives us x = 0. Similarly, the partial derivative with respect to y is f_y(x, y) = -16y, and setting this equal to zero gives y = 0. Therefore, the critical point is at (0, 0).
To classify the nature of this critical point using the second derivative test, we calculate the second partial derivatives: f_xx(x, y) = -14, f_yy(x, y) = -16, and the mixed partial derivative f_xy(x, y) = 0. The test involves evaluating the determinant D = f_xx * f_yy - (f_xy)^2 at the critical point. At (0, 0), we get D = (-14) * (-16) - (0)^2 = 224, which is positive. Since both f_xx(0, 0) and f_yy(0, 0) are negative, the critical point (0, 0) corresponds to a local maximum.