Final answer:
The linearization L(x,y) of the function f(x,y) at (-3,4) is found by calculating the gradient at that point and then using it to form the equation of the tangent plane, resulting in L(x,y) = -3x + 4y + √{338}.
Step-by-step explanation:
To find the linearization L(x,y) of the function f(x,y) = √{338 - 9x^2 - 16y^2} at the point (-3,4), we need to first calculate the gradient of f at that point. The gradient is given by the vector of partial derivatives of the function, which can be written as ∇f = (f_x, f_y), where f_x is the partial derivative with respect to x, and f_y is the partial derivative with respect to y.
At the point (-3,4), the function value is √{338 - 9(-3)^2 - 16(4)^2} = √{338 - 81 - 256} = √{1} = 1. The partial derivatives are f_x = -9x/√{338 - 9x^2 - 16y^2} and f_y = -16y/√{338 - 9x^2 - 16y^2}. Thus at the point (-3,4), we have f_x(-3,4) = 9(3)/1 = 27 and f_y(-3,4) = 16(4)/1 = 64.
The linearization L(x,y) of the function at the point (-3,4) is then L(x,y) = f(-3,4) + f_x(-3,4)(x+3) + f_y(-3,4)(y-4), which simplifies to L(x,y) = 1 + 27(x+3) + 64(y-4) or L(x,y) = -3x + 4y + √{338} when we rearrange the terms. Therefore, the correct answer is B. L(x,y) = -3x + 4y + √{338}.