1. **State the problem:** Find the minimum and maximum values of the function $$f(x,y) = x^2 - xy + y^2$$ on the quarter circle defined by $$x^2 + y^2 = 1$$ with $$x, y \geq 0$$. 2. **Use the constraint:** Since $$x^2 + y^2 = 1$$, we can parametrize the quarter circle using $$x = \cos \theta$$ and $$y = \sin \theta$$ where $$\theta \in [0, \frac{\pi}{2}]$$ because $$x,y \geq 0$$. 3. **Rewrite the function in terms of $$\theta$$:** $$$ f(\theta) = (\cos \theta)^2 - (\cos \theta)(\sin \theta) + (\sin \theta)^2 = \cos^2 \theta - \cos \theta \sin \theta + \sin^2 \theta $$$ 4. **Simplify using the Pythagorean identity $$\cos^2 \theta + \sin^2 \theta = 1$$:** $$$ f(\theta) = 1 - \cos \theta \sin \theta $$$ 5. **Find critical points by differentiating $$f(\theta)$$ with respect to $$\theta$$:** $$$ f'(\theta) = - ( -\sin \theta \sin \theta + \cos \theta \cos \theta ) = - ( -\sin^2 \theta + \cos^2 \theta ) = - ( \cos^2 \theta - \sin^2 \theta ) $$$ 6. **Set derivative to zero to find extrema:** $$$ f'(\theta) = 0 \implies \cos^2 \theta - \sin^2 \theta = 0 \implies \cos^2 \theta = \sin^2 \theta $$$ 7. **Solve for $$\theta$$:** $$$ \cos^2 \theta = \sin^2 \theta \implies \tan^2 \theta = 1 \implies \tan \theta = \pm 1 $$$ Since $$\theta \in [0, \frac{\pi}{2}]$$, only $$\tan \theta = 1$$ is valid, so $$$ \theta = \frac{\pi}{4} $$$ 8. **Evaluate $$f(\theta)$$ at critical point and boundaries:** - At $$\theta = 0$$: $$$ f(0) = 1 - \cos 0 \sin 0 = 1 - 1 \times 0 = 1 $$$ - At $$\theta = \frac{\pi}{4}$$: $$$ f\left(\frac{\pi}{4}\right) = 1 - \cos \frac{\pi}{4} \sin \frac{\pi}{4} = 1 - \frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2} = 1 - \frac{1}{2} = \frac{1}{2} $$$ - At $$\theta = \frac{\pi}{2}$$: $$$ f\left(\frac{\pi}{2}\right) = 1 - \cos \frac{\pi}{2} \sin \frac{\pi}{2} = 1 - 0 \times 1 = 1 $$$ 9. **Conclusion:** The minimum value of $$f(x,y)$$ on the quarter circle is $$\frac{1}{2}$$ at $$\theta = \frac{\pi}{4}$$, which corresponds to $$x = y = \frac{\sqrt{2}}{2}$$. The maximum value is $$1$$ at the points $$\theta = 0$$ (i.e., $$x=1,y=0$$) and $$\theta = \frac{\pi}{2}$$ (i.e., $$x=0,y=1$$). **Final answer:** $$$ \min f = \frac{1}{2} \text{ at } \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right), \quad \max f = 1 \text{ at } (1,0) \text{ and } (0,1) $$$