1. **State the problem:** Find all critical points (minima and maxima) of the function $$f(n,y) = n^4 + y^4 - 4ny + 1$$ by solving where the partial derivatives are zero. 2. **Find partial derivatives:** $$\frac{\partial f}{\partial n} = 4n^3 - 4y$$ $$\frac{\partial f}{\partial y} = 4y^3 - 4n$$ 3. **Set the partial derivatives equal to zero to find critical points:** $$4n^3 - 4y = 0 \implies y = n^3$$ $$4y^3 - 4n = 0 \implies y^3 = n$$ 4. **Substitute $y = n^3$ into $y^3 = n$:** $$ (n^3)^3 = n \implies n^9 = n$$ 5. **Solve for $n$:** $$n^9 - n = 0$$ $$n(n^8 - 1) = 0$$ Hence, $$n = 0$$ or $$n^8 = 1 \implies n = \pm 1$$ 6. **Find corresponding $y$ values:** - For $$n=0$$: $$y = 0^3 = 0$$ - For $$n=1$$: $$y = 1^3 = 1$$ - For $$n=-1$$: $$y = (-1)^3 = -1$$ 7. **Critical points are:** $$(0,0), (1,1), (-1,-1)$$ 8. **Classify critical points using the second derivative test:** Calculate second derivatives: $$f_{nn} = 12n^2$$ $$f_{yy} = 12y^2$$ $$f_{ny} = f_{yn} = -4$$ Evaluate discriminant $$D = f_{nn}f_{yy} - (f_{ny})^2$$ at each critical point: - At $$(0,0)$$: $$f_{nn} = 0, f_{yy} = 0, f_{ny} = -4$$ $$D = 0 \cdot 0 - (-4)^2 = -16 < 0$$ Thus, $$(0,0)$$ is a saddle point. - At $$(1,1)$$: $$f_{nn} = 12(1)^2 = 12, f_{yy} = 12(1)^2 = 12, f_{ny} = -4$$ $$D = 12 \cdot 12 - 16 = 144 - 16 = 128 > 0$$ Since $$f_{nn} = 12 > 0$$, $$(1,1)$$ is a local minimum. - At $$(-1,-1)$$: $$f_{nn} = 12(-1)^2 = 12, f_{yy} = 12(-1)^2 = 12, f_{ny} = -4$$ $$D = 12 \cdot 12 - 16 = 128 > 0$$ Since $$f_{nn} = 12 > 0$$, $$(-1,-1)$$ is also a local minimum. **Final answer:** - Saddle point at $$(0,0)$$ - Local minima at $$(1,1)$$ and $$(-1,-1)$$