We are tasked with finding the linear approximation of the function $f(x, y) = 4\sqrt{\frac{xy}{2}}$ at the point $(4, 2)$, and then using it to approximate $f(4.22, 2.13)$.

Step-by-step solution:

1. Function and Point of Interest: The given function is: $f(x, y) = 4\sqrt{\frac{xy}{2}}$ The point where we approximate is $(4, 2)$.

2. Find the partial derivatives: To create the linear approximation, we need the partial derivatives $f_x$ and $f_y$ at the point $(4, 2)$.

   • First, compute the partial derivative with respect to $x$: = 4 \cdot \frac{1}{2} \cdot \left( \frac{y}{2} \right)^{-\frac{1}{2}} \cdot \frac{y}{2} = \frac{4y}{2\sqrt{2xy}}$$
   • Now, compute the partial derivative with respect to $y$: = 4 \cdot \frac{1}{2} \cdot \left( \frac{xy}{2} \right)^{-\frac{1}{2}} \cdot \frac{x}{2} = \frac{4x}{2\sqrt{2xy}}$$

3. Evaluate the function and partial derivatives at $(4, 2)$:

   • Evaluate the function $f(x, y)$ at $(4, 2)$: $f(4, 2) = 4\sqrt{\frac{4 \cdot 2}{2}} = 4\sqrt{4} = 4 \times 2 = 8$

   • Evaluate $f_x(4, 2)$: $f_x(4, 2) = \frac{4 \times 2}{2\sqrt{2 \times 4 \times 2}} = \frac{8}{2\sqrt{16}} = \frac{8}{8} = 1$

   • Evaluate $f_y(4, 2)$: $f_y(4, 2) = \frac{4 \times 4}{2\sqrt{2 \times 4 \times 2}} = \frac{16}{8} = 2$

4. Linear approximation formula: The linear approximation formula is: $L(x, y) = f(4, 2) + f_x(4, 2)(x - 4) + f_y(4, 2)(y - 2)$ Substituting the values we found: $L(x, y) = 8 + 1(x - 4) + 2(y - 2)$ Simplify the expression:

   = 8 + x - 4 + 2y - 4 = x + 2y$$

5. Approximate $f(4.22, 2.13)$: Now, we use the linear approximation to approximate $f(4.22, 2.13)$: $f(4.22, 2.13) \approx L(4.22, 2.13) = 4.22 + 2 \times 2.13 = 4.22 + 4.26 = 8.48$

Thus, the approximate value of $f(4.22, 2.13)$ is 8.480 (rounded to three decimal places).

Would you like more details or have any other questions?

Here are five related questions to explore:

1. How does the method of linear approximation work in general?
2. Why are partial derivatives crucial in calculating linear approximations?
3. Can you approximate $f(4.1, 1.9)$ using the same function?
4. How does the linear approximation compare to the actual value of the function at nearby points?
5. How would this method change if the function were in three variables, say $f(x, y, z)$?

Tip: Linear approximations are most accurate for points very close to the given point of approximation, so the closer your point, the more precise the approximation!