To solve for the absolute maximum and minimum values of $f(x, y) = 2x - 2y$ subject to the constraint $g(x, y) = x^2 - xy + y^2 - 12 = 0$, we use the method of Lagrange multipliers.

The method involves solving the equations:

$\nabla f = \lambda \nabla g$

where:

• $\nabla f = \langle 2, -2 \rangle$
• $\nabla g = \langle 2x - y, -x + 2y \rangle$.

Steps to solve:

1. Set up the Lagrange equations: 2 = \lambda (2x - y) \tag{1} -2 = \lambda (-x + 2y) \tag{2} x^2 - xy + y^2 - 12 = 0 \tag{3 (constraint)}

2. Solve for $\lambda$ from the first two equations:

   From (1): $\lambda = \frac{2}{2x - y}$ (provided $2x - y \neq 0$).

   From (2): $\lambda = \frac{-2}{-x + 2y} = \frac{2}{x - 2y}$.

   Equating the two expressions for $\lambda$: $\frac{2}{2x - y} = \frac{2}{x - 2y}$

   Simplify: $2x - y = x - 2y$

   Solve for $x$ and $y$: $x = y$

3. Substitute $x = y$ into the constraint:

   Replace $x = y$ in $x^2 - xy + y^2 - 12 = 0$: $x^2 - x(x) + x^2 - 12 = 0$ $2x^2 - 12 = 0$ $x^2 = 6 \quad \implies \quad x = \pm \sqrt{6}$

   Since $x = y$, $y = \pm \sqrt{6}$.

4. Substitute $x = \pm \sqrt{6}, y = \pm \sqrt{6}$ into $f(x, y)$:

   For $x = \sqrt{6}, y = \sqrt{6}$: $f(x, y) = 2x - 2y = 2\sqrt{6} - 2\sqrt{6} = 0$

   For $x = -\sqrt{6}, y = -\sqrt{6}$: $f(x, y) = 2x - 2y = 2(-\sqrt{6}) - 2(-\sqrt{6}) = 0$

   Both points yield $f(x, y) = 0$.

Additional critical points (check $2x - y = 0$ or $x - 2y = 0$):

If $2x - y = 0$: $y = 2x$ Substitute into the constraint $x^2 - xy + y^2 = 12$: $x^2 - x(2x) + (2x)^2 = 12$ $x^2 - 2x^2 + 4x^2 = 12$ $3x^2 = 12 \quad \implies \quad x^2 = 4 \quad \implies \quad x = \pm 2$

If $x = 2$, $y = 4$. If $x = -2$, $y = -4$.

If $x - 2y = 0$: $x = 2y$ Substitute into the constraint $x^2 - xy + y^2 = 12$: $(2y)^2 - (2y)y + y^2 = 12$ $4y^2 - 2y^2 + y^2 = 12$ $3y^2 = 12 \quad \implies \quad y^2 = 4 \quad \implies \quad y = \pm 2$

If $y = 2$, $x = 4$. If $y = -2$, $x = -4$.

Evaluate $f(x, y)$ at these points:

1. $(2, 4)$: $f(x, y) = 2(2) - 2(4) = 4 - 8 = -4$.
2. $(-2, -4)$: $f(x, y) = 2(-2) - 2(-4) = -4 + 8 = 4$.

Final Answer:

• Absolute maximum value: $4$ at $(-2, -4)$.
• Absolute minimum value: $-4$ at $(2, 4)$.

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Would you like a detailed explanation of any step or the general method? 😊

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Related Questions:

1. What are the key applications of Lagrange multipliers in optimization?
2. How do you handle constraints that include inequalities instead of equations?
3. Can the method of Lagrange multipliers work for functions with three or more variables?
4. How would a second constraint alter the optimization problem?
5. What are the geometric interpretations of Lagrange multipliers?

Tip:

Always verify critical points by substituting them back into the original constraint equation to ensure they are valid.