1. **Stating the problem:** We need to find the conditional extrema (maximum or minimum) of the function $$f(x,y) = 2x^2 - xy$$ subject to the constraint $$u(x,y) = 2x + 2y - 2 = 0$$. 2. **Method:** We use the method of Lagrange multipliers. The condition for extrema under constraint is given by: $$\nabla f(x,y) = \lambda \nabla u(x,y)$$ where $$\lambda$$ is the Lagrange multiplier. 3. **Calculate gradients:** $$\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (4x - y, -x)$$ $$\nabla u = \left( \frac{\partial u}{\partial x}, \frac{\partial u}{\partial y} \right) = (2, 2)$$ 4. **Set up the system:** $$4x - y = 2\lambda$$ $$-x = 2\lambda$$ $$2x + 2y - 2 = 0$$ 5. **From the second equation:** $$-x = 2\lambda \implies \lambda = -\frac{x}{2}$$ 6. **Substitute $$\lambda$$ into the first equation:** $$4x - y = 2 \left(-\frac{x}{2}\right) = -x$$ 7. **Simplify:** $$4x - y = -x \implies 4x + x = y \implies y = 5x$$ 8. **Use the constraint:** $$2x + 2y - 2 = 0 \implies 2x + 2(5x) - 2 = 0$$ $$2x + 10x - 2 = 0 \implies 12x = 2 \implies x = \frac{1}{6}$$ 9. **Find $$y$$:** $$y = 5x = 5 \times \frac{1}{6} = \frac{5}{6}$$ 10. **Find $$\lambda$$:** $$\lambda = -\frac{x}{2} = -\frac{1/6}{2} = -\frac{1}{12}$$ 11. **Evaluate $$f$$ at $$(x,y) = \left(\frac{1}{6}, \frac{5}{6}\right)$$:** $$f\left(\frac{1}{6}, \frac{5}{6}\right) = 2 \left(\frac{1}{6}\right)^2 - \left(\frac{1}{6}\right) \left(\frac{5}{6}\right) = 2 \times \frac{1}{36} - \frac{5}{36} = \frac{2}{36} - \frac{5}{36} = -\frac{3}{36} = -\frac{1}{12}$$ 12. **Conclusion:** The function has a conditional extremum at $$\left(\frac{1}{6}, \frac{5}{6}\right)$$ with value $$f = -\frac{1}{12}$$ under the constraint $$2x + 2y - 2 = 0$$. **Final answer:** $$\boxed{\left(x,y\right) = \left(\frac{1}{6}, \frac{5}{6}\right), \quad f = -\frac{1}{12}}$$