Find the absolute extrema of the given function on the indicated closed and bounded set R. (Order your answers from smallest to largest x, then from smallest to largest y.)
f(x, y) = x3 − 6xy − y3  on  R = {(x, y): −3 ≤ x ≤ 3, −3 ≤ y ≤ 3}

To find the absolute extrema of the function \( f(x, y) = x^3 - 6xy - y^3 \) on the closed and bounded region \( R = \{(x, y): -3 \leq x \leq 3, -3 \leq y \leq 3 \} \), we follow these steps:

### 1. **Find the critical points in the interior of \( R \)**

We first compute the partial derivatives of \( f(x, y) \) and set them equal to zero to find the critical points.

#### a) Compute the partial derivatives:

\[
f_x(x, y) = \frac{\partial}{\partial x}(x^3 - 6xy - y^3) = 3x^2 - 6y
\]
\[
f_y(x, y) = \frac{\partial}{\partial y}(x^3 - 6xy - y^3) = -6x - 3y^2
\]

#### b) Set the partial derivatives equal to zero to find the critical points:

\[
3x^2 - 6y = 0 \quad \text{(1)}
\]
\[
-6x - 3y^2 = 0 \quad \text{(2)}
\]

From equation (1), solve for \( y \):

\[
3x^2 - 6y = 0 \implies y = \frac{x^2}{2} \quad \text{(3)}
\]

Substitute equation (3) into equation (2):

\[
-6x - 3\left(\frac{x^2}{2}\right)^2 = 0
\]

This simplifies to:

\[
-6x - 3 \left(\frac{x^4.append(String)}} \ti}




Find the absolute extrema of the given function on the indicated closed and bounded set R. (Order your answers from smallest to largest x, then from smallest to largest y.) f(x, y) = x^3 − 6xy − y^3 on R = {(x, y): −3 ≤ x ≤ 3, −3 ≤ y ≤ 3}

Learn how to find the absolute extrema of the function f(x, y) = x^3 - 6xy - y^3 on the region R = {(x, y): −3 ≤ x ≤ 3, −3 ≤ y ≤ 3}. This problem uses multivariable calculus, optimization techniques, and partial derivatives to find critical points and determine extrema on a closed and bounded set. Suitable for college students studying Calculus III or multivariable calculus.

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