1. **State the problem:** We have the function $$f(x,y) = (x^2 + 2xy)e^y$$ defined on $$\mathbb{R}^2$$. We need to find the partial derivatives $$\frac{\partial f}{\partial x}$$ and $$\frac{\partial f}{\partial y}$$, then find and classify all critical points (minimum, maximum, saddle). 2. **Find the partial derivatives:** - For $$\frac{\partial f}{\partial x}$$, treat $$y$$ as constant: $$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left((x^2 + 2xy)e^y\right) = e^y \frac{\partial}{\partial x} (x^2 + 2xy) = e^y (2x + 2y) = 2e^y (x + y)$$ - For $$\frac{\partial f}{\partial y}$$, treat $$x$$ as constant and use product rule: $$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left((x^2 + 2xy)e^y\right) = (x^2 + 2xy) \frac{\partial}{\partial y} e^y + e^y \frac{\partial}{\partial y} (x^2 + 2xy) = (x^2 + 2xy)e^y + e^y (0 + 2x) = e^y (x^2 + 2xy + 2x)$$ 3. **Find critical points:** Critical points satisfy: $$\frac{\partial f}{\partial x} = 0 \quad \Rightarrow \quad 2e^y (x + y) = 0$$ Since $$e^y \neq 0$$ for all $$y$$, this implies: $$x + y = 0 \quad \Rightarrow \quad y = -x$$ Similarly, $$\frac{\partial f}{\partial y} = 0 \quad \Rightarrow \quad e^y (x^2 + 2xy + 2x) = 0$$ Again, $$e^y \neq 0$$, so: $$x^2 + 2xy + 2x = 0$$ Substitute $$y = -x$$: $$x^2 + 2x(-x) + 2x = x^2 - 2x^2 + 2x = -x^2 + 2x = 0$$ 4. **Solve for $$x$$:** $$-x^2 + 2x = 0 \Rightarrow x(-x + 2) = 0$$ So, $$x = 0 \quad \text{or} \quad x = 2$$ 5. **Find corresponding $$y$$ values:** - If $$x=0$$, then $$y = -0 = 0$$ - If $$x=2$$, then $$y = -2$$ 6. **Critical points:** $$(0,0) \quad \text{and} \quad (2,-2)$$ 7. **Classify critical points using second derivative test:** Calculate second partial derivatives: $$f_{xx} = \frac{\partial}{\partial x} \left(2e^y (x + y)\right) = 2e^y \frac{\partial}{\partial x} (x + y) = 2e^y (1 + 0) = 2e^y$$ $$f_{yy} = \frac{\partial}{\partial y} \left(e^y (x^2 + 2xy + 2x)\right) = e^y (x^2 + 2xy + 2x) + e^y \frac{\partial}{\partial y} (x^2 + 2xy + 2x) = e^y (x^2 + 2xy + 2x) + e^y (0 + 2x + 0) = e^y (x^2 + 2xy + 2x + 2x) = e^y (x^2 + 2xy + 4x)$$ $$f_{xy} = \frac{\partial}{\partial y} \left(2e^y (x + y)\right) = 2e^y (x + y) + 2e^y (0 + 1) = 2e^y (x + y + 1)$$ 8. **Evaluate second derivatives at critical points:** - At $$(0,0)$$: $$f_{xx} = 2e^0 = 2$$ $$f_{yy} = e^0 (0 + 0 + 0) = 0$$ $$f_{xy} = 2e^0 (0 + 0 + 1) = 2$$ Discriminant: $$D = f_{xx} f_{yy} - (f_{xy})^2 = 2 \times 0 - 2^2 = -4 < 0$$ Since $$D < 0$$, $$(0,0)$$ is a saddle point. - At $$(2,-2)$$: Calculate each term: $$e^{-2} > 0$$ $$f_{xx} = 2e^{-2}$$ $$f_{yy} = e^{-2} (2^2 + 2 \times 2 \times (-2) + 4 \times 2) = e^{-2} (4 - 8 + 8) = e^{-2} (4) = 4e^{-2}$$ $$f_{xy} = 2e^{-2} (2 + (-2) + 1) = 2e^{-2} (1) = 2e^{-2}$$ Discriminant: $$D = f_{xx} f_{yy} - (f_{xy})^2 = (2e^{-2})(4e^{-2}) - (2e^{-2})^2 = 8e^{-4} - 4e^{-4} = 4e^{-4} > 0$$ Since $$D > 0$$ and $$f_{xx} > 0$$, $$(2,-2)$$ is a local minimum. **Final answers:** - $$\frac{\partial f}{\partial x} = 2e^y (x + y)$$ - $$\frac{\partial f}{\partial y} = e^y (x^2 + 2xy + 2x)$$ - Critical points and classification: - $$(0,0)$$ saddle point - $$(2,-2)$$ local minimum