1. **Problem statement:** Given the function $$f(x,y) = \frac{1}{3}x^3 + \frac{4}{3} y^3 - x^2 - 3x - 4y - 2025,$$ we need to find partial derivatives, critical points, the discriminant, and classify the critical points using the second derivative test. 2. **Find partial derivatives:** - The partial derivative with respect to $x$ is $$f_x = \frac{\partial}{\partial x} \left( \frac{1}{3}x^3 + \frac{4}{3} y^3 - x^2 - 3x - 4y - 2025 \right) = x^2 - 2x - 3.$$ - The mixed partial derivative $$f_{xy} = \frac{\partial}{\partial y} f_x = \frac{\partial}{\partial y} (x^2 - 2x - 3) = 0,$$ since $x^2 - 2x - 3$ has no $y$ term. - The second partial derivative with respect to $x$ is $$f_{xx} = \frac{\partial}{\partial x} f_x = \frac{\partial}{\partial x} (x^2 - 2x - 3) = 2x - 2.$$ 3. **Find partial derivatives with respect to $y$:** - The partial derivative with respect to $y$ is $$f_y = \frac{\partial}{\partial y} \left( \frac{1}{3}x^3 + \frac{4}{3} y^3 - x^2 - 3x - 4y - 2025 \right) = 4y^2 - 4.$$ - The second partial derivative with respect to $y$ is $$f_{yy} = \frac{\partial}{\partial y} f_y = \frac{\partial}{\partial y} (4y^2 - 4) = 8y.$$ 4. **Find critical points:** - Critical points occur where $$f_x = 0$$ and $$f_y = 0.$$ - Solve $$x^2 - 2x - 3 = 0$$ which factors as $$(x - 3)(x + 1) = 0,$$ so $$x = 3$$ or $$x = -1.$$ - Solve $$4y^2 - 4 = 0$$ which simplifies to $$y^2 = 1,$$ so $$y = \pm 1.$$ - Therefore, critical points are $$(3,1), (3,-1), (-1,1), (-1,-1).$$ 5. **Discriminant function $$\triangle$$:** - Given $$f_{xy} = f_{yx} = 0,$$ the discriminant is $$\triangle = f_{xx} f_{yy} - (f_{xy})^2 = (2x - 2)(8y) - 0 = 8y(2x - 2).$$ 6. **Second derivative test classification:** - For each critical point, compute $$\triangle$$ and $$f_{xx}$$: - At $$(3,1):$$ $$\triangle = 8(1)(2(3) - 2) = 8 \times 1 \times 4 = 32 > 0,$$ and $$f_{xx} = 2(3) - 2 = 4 > 0,$$ so this is a local minimum. - At $$(3,-1):$$ $$\triangle = 8(-1)(4) = -32 < 0,$$ so this is a saddle point. - At $$(-1,1):$$ $$\triangle = 8(1)(2(-1) - 2) = 8 \times 1 \times (-4) = -32 < 0,$$ so this is a saddle point. - At $$(-1,-1):$$ $$\triangle = 8(-1)(-4) = 32 > 0,$$ and $$f_{xx} = 2(-1) - 2 = -4 < 0,$$ so this is a local maximum. **Final answers:** - $$f_x = x^2 - 2x - 3,$$ - $$f_{xy} = 0,$$ - $$f_{xx} = 2x - 2,$$ - $$f_y = 4y^2 - 4,$$ - $$f_{yy} = 8y,$$ - Critical points: $$(3,1), (3,-1), (-1,1), (-1,-1),$$ - Discriminant: $$\triangle = 8y(2x - 2),$$ - Classification: local minimum at $$(3,1),$$ saddle points at $$(3,-1)$$ and $$(-1,1),$$ local maximum at $$(-1,-1).$$