Final answer:
The linear approximation of the function f(x, y) at the point (5, 1) is 9 - 10(x - 5) - 8(y - 1). Plugging in (5.06, 1.06) into the approximation provides an estimated value of 7.92 for f(5.06, 1.06).
Step-by-step explanation:
To find the linear approximation of the function f(x, y) = 38 − x2 − 4y2 at the point (5, 1), we need to determine the gradient of f at that point and use it to approximate the function's value near (5, 1). The gradient is a vector composed of partial derivatives of f with respect to x and y. Hence, ∂f/∂x = −2x and ∂f/∂y = −8y.
At the point (5, 1), the gradient ∇f is (-10, -8). The equation for the linear approximation, or the tangent plane, at a point (x0, y0) is given by:
f(x, y) ≈ f(x0, y0) + ∇f · (x − x0, y − y0)
Substituting the given values, we get:
f(5, 1) ≈ 38 − 52 − 4(12) + (-10)(x − 5) + (-8)(y − 1)
Simplifying, we find:
f(5, 1) ≈ 38 − 25 − 4 + (-10)(x − 5) + (-8)(y − 1) = 9 − 10(x − 5) − 8(y − 1)
Now, to approximate f(5.06, 1.06), we plug in x = 5.06 and y = 1.06 into our linear approximation formula:
f(5.06, 1.06) ≈ 9 − 10(5.06 − 5) − 8(1.06 − 1) = 9 − 10(0.06) − 8(0.06) = 9 − 0.6 − 0.48 = 7.92
The approximated value of f(5.06, 1.06), rounded to three decimal places, is 7.920.