Lagrange Circle 929339

1. **State the problem:** We want to analyze the function $$f(x,y) = 2x^3 - 9x^2 - 24x + y^2 - 2y + 1$$ subject to the constraint that $(x,y)$ lies on the circle $$C: x^2 + y^2 = 4$$. 2. **Identify the goal:** Typically, such problems ask to find the extrema (maximum and minimum values) of $f$ on the circle $C$. 3. **Method:** We use the method of Lagrange multipliers to find critical points of $f$ subject to the constraint $g(x,y) = x^2 + y^2 - 4 = 0$. 4. **Set up Lagrange system:** We introduce a multiplier $\lambda$ and solve: $$\nabla f = \lambda \nabla g$$ which means: $$\frac{\partial f}{\partial x} = \lambda \frac{\partial g}{\partial x}, \quad \frac{\partial f}{\partial y} = \lambda \frac{\partial g}{\partial y}$$ 5. **Calculate partial derivatives:** $$\frac{\partial f}{\partial x} = 6x^2 - 18x - 24$$ $$\frac{\partial f}{\partial y} = 2y - 2$$ $$\frac{\partial g}{\partial x} = 2x$$ $$\frac{\partial g}{\partial y} = 2y$$ 6. **Write the system:** $$6x^2 - 18x - 24 = \lambda 2x$$ $$2y - 2 = \lambda 2y$$ $$x^2 + y^2 = 4$$ 7. **Analyze the $y$-equation:** $$2y - 2 = 2\lambda y \implies 2y - 2 = 2\lambda y$$ Rearranged: $$2y - 2\lambda y = 2 \implies 2y(1 - \lambda) = 2 \implies y(1 - \lambda) = 1$$ Two cases: - If $y \neq 0$, then $\lambda = 1 - \frac{1}{y}$ - If $y = 0$, then from the equation $2y - 2 = 2\lambda y$ we get $-2 = 0$, which is false, so $y=0$ is not possible. 8. **Substitute $\lambda$ into $x$-equation:** $$6x^2 - 18x - 24 = 2x \lambda = 2x \left(1 - \frac{1}{y}\right) = 2x - \frac{2x}{y}$$ Rearranged: $$6x^2 - 18x - 24 - 2x + \frac{2x}{y} = 0$$ $$6x^2 - 20x - 24 + \frac{2x}{y} = 0$$ 9. **Use the constraint $x^2 + y^2 = 4$ to express $x$ or $y$ and solve the system:** This is a nonlinear system that can be solved numerically or by substitution. 10. **Summary:** - Use $y(1 - \lambda) = 1$ to express $\lambda$ in terms of $y$. - Substitute into $x$-equation. - Use $x^2 + y^2 = 4$. - Solve for $(x,y)$. 11. **Evaluate $f(x,y)$ at each solution to find extrema on $C$.