Final answer:
The linearization of f(x, y) = x^2 - xy + y^2/2 + 3 at the point (3, 2) can be found using the equation for linearization. The partial derivatives are evaluated at (3, 2) and substituted into the linearization equation.
Step-by-step explanation:
The linearization of f(x, y) = x^2 - xy + y^2/2 + 3 at the point (3, 2) can be found using the equation for linearization: L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b), where f_x(a, b) and f_y(a, b) are the partial derivatives of f(x, y) with respect to x and y evaluated at (a, b).
First, we find the partial derivatives: f_x(x, y) = 2x - y, and f_y(x, y) = -x + y/2.
Then, we evaluate the partial derivatives at (3, 2): f_x(3, 2) = 2(3) - 2 = 4, and f_y(3, 2) = -3 + 2/2 = -2.
Finally, we substitute these values into the linearization equation: L(x, y) = (3^2 - 3(2) + (2^2)/2 + 3) + 4(x - 3) - 2(y - 2).