Find the quadratic approximation to
f(x, y) = p
1 − 4x − 2y
at P(0, 0).

The quadratic approximation of a multivariable function \( f(x, y) \) at a point \( P(x_0, y_0) \) is given by the Taylor expansion up to the second-order terms around the point.

For a function \( f(x, y) \), the quadratic approximation around the point \( (x_0, y_0) \) is:

\[
f(x, y) \approx f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) + \frac{1}{2} f_{xx}(x_0, y_0)(x - x_0)^2 + f_{xy}(x_0, y_0)(x - x_0)(y - y_0) + \frac{1}{2} f_{yy}(x_0, y_0)(y - y_0)^2
\]
where:
- \( f_x \), \( f_y \) are the first partial derivatives, and
- \( f_{xx} \), \( f_{xy} \), \( f_{yy} \) are the second partial derivatives.

Now, let's calculate the quadratic approximation for \( f(x, y) = \frac{p}{1 - 4x - 2y} \) at the point \( (0, 0) \).

### Step 1: Evaluate \( f(0, 0) \)
At \( (0, 0) \), the function is:
\[
f(0, 0) = \frac{p}{1 - 4(0) - 2(0)} = \frac{p}{1} = p
\]

### Step 2: Find the first partial derivatives
The first partial derivatives are:
\[
f_x(x, y) = \frac{\partial}{\partial x} \left( \frac{p}{1 - 4x - 2y} \right) = \frac{4p}{(1 - 4x - 2y)^2}
\]
\[
f_y(x, y) = \frac{\partial}{\partial y} \left( \frac{p}{1 - 4x - 2y} \right) = \frac{2p}{(1 - 4x - 2y)^2}
\]

Evaluating at \( (0, 0) \):
\[
f_x(0, 0) = \frac{4p}{(1)^2} = 4p
\]
\[
f_y(0, 0) = \frac{2p}{(1)^2} = 2p
\]

### Step 3: Find the second partial derivatives
The second partial derivatives are:
\[
f_{xx}(x, y) = \frac{\partial}{\partial x} \left( \frac{4p}{(1 - 4x - 2y)^2} \right) = \frac{32p}{(1 - 4x - 2y)^3}
\]
\[
f_{yy}(x, y) = \frac{\partial}{\partial y} \left( \frac{2p}{(1 - 4x - 2y)^2} \right) = \frac{8p}{(1 - 4x - 2y)^3}
\]
\[
f_{xy}(x, y) = \frac{\partial}{\partial y} \left( \frac{4p}{(1 - 4x - 2y)^2} \right) = \frac{16p}{(1 - 4x - 2y)^3}
\]

Evaluating at \( (0, 0) \):
\[
f_{xx}(0, 0) = \frac{32p}{(1)^3} = 32p
\]
\[
f_{yy}(0, 0) = \frac{8p}{(1)^3} = 8p
\]
\[
f_{xy}(0, 0) = \frac{16p}{(1)^3} = 16p
\]

### Step 4: Write the quadratic approximation
Using the Taylor series formula, we have the quadratic approximation at \( (0, 0) \):
\[
f(x, y) \approx p + 4p \cdot x + 2p \cdot y + \frac{1}{2} \cdot 32p \cdot x^2 + 16p \cdot xy + \frac{1}{2} \cdot 8p \cdot y^2
\]

Simplifying:
\[
f(x, y) \approx p + 4p x + 2p y + 16p x^2 + 16p xy + 4p y^2
\]

This is the quadratic approximation of the function \( f(x, y) = \frac{p}{1 - 4x - 2y} \) at the point \( (0, 0) \).

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Would you like further clarification on any part of this solution? Here are five related questions to deepen your understanding:

1. How does the quadratic approximation change if the point of approximation is different from \( (0, 0) \)?
2. What are the implications of second-order partial derivatives in Taylor expansions?
3. Can you derive the quadratic approximation for \( f(x, y) = \ln(1 + 4x + 2y) \)?
4. How is the quadratic approximation used in optimization problems?
5. What would happen if higher-order terms were included in the approximation?

**Tip**: The quadratic approximation is often called the second-order Taylor expansion, and it's particularly useful for approximating nonlinear functions near a point.

Find the quadratic approximation to f(x, y) = p / (1 - 4x - 2y) at P(0, 0).

Explore the quadratic approximation of the multivariable function f(x, y) = p / (1 - 4x - 2y) at the point (0, 0). This step-by-step solution includes the first and second partial derivatives and uses Taylor Series Expansion. Ideal for students in Calculus III.

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