1. **State the problem:** Find and classify the extrema of the function $F(x,y) = 2x^2 + 10xy + 3y^2$ on the boundary of the circle $x^2 + y^2 = 5$. 2. **Method:** Use Lagrange multipliers with constraint $g(x,y) = x^2 + y^2 - 5 = 0$. 3. **Set up Lagrangian:** $$\mathcal{L}(x,y,\lambda) = 2x^2 + 10xy + 3y^2 + \lambda (x^2 + y^2 - 5)$$ 4. **Find partial derivatives:** $$\frac{\partial \mathcal{L}}{\partial x} = 4x + 10y + 2\lambda x = 0$$ $$\frac{\partial \mathcal{L}}{\partial y} = 10x + 6y + 2\lambda y = 0$$ $$\frac{\partial \mathcal{L}}{\partial \lambda} = x^2 + y^2 - 5 = 0$$ 5. **Rewrite system:** $$4x + 10y + 2\lambda x = 0$$ $$10x + 6y + 2\lambda y = 0$$ $$x^2 + y^2 = 5$$ 6. **Express as matrix system:** $$\begin{pmatrix}4 + 2\lambda & 10 \\ 10 & 6 + 2\lambda \end{pmatrix} \begin{pmatrix}x \\ y \end{pmatrix} = \begin{pmatrix}0 \\ 0 \end{pmatrix}$$ 7. **Nontrivial solution requires determinant zero:** $$\det \begin{pmatrix}4 + 2\lambda & 10 \\ 10 & 6 + 2\lambda \end{pmatrix} = 0$$ Calculate determinant: $$ (4 + 2\lambda)(6 + 2\lambda) - 100 = 0 $$ $$ 24 + 8\lambda + 12\lambda + 4\lambda^2 - 100 = 0 $$ $$ 4\lambda^2 + 20\lambda - 76 = 0 $$ 8. **Simplify quadratic:** $$ \lambda^2 + 5\lambda - 19 = 0 $$ 9. **Solve quadratic:** $$ \lambda = \frac{-5 \pm \sqrt{25 + 76}}{2} = \frac{-5 \pm \sqrt{101}}{2} $$ 10. **Find eigenvectors for each $\lambda$:** For $\lambda_1 = \frac{-5 + \sqrt{101}}{2}$, solve $$ (4 + 2\lambda_1)x + 10y = 0 $$ For $\lambda_2 = \frac{-5 - \sqrt{101}}{2}$, solve $$ (4 + 2\lambda_2)x + 10y = 0 $$ 11. **Express $y$ in terms of $x$ for each $\lambda$:** $$ y = -\frac{4 + 2\lambda}{10} x $$ 12. **Use constraint $x^2 + y^2 = 5$ to find $x,y$:** $$ x^2 + \left(-\frac{4 + 2\lambda}{10} x\right)^2 = 5 $$ $$ x^2 \left(1 + \left(\frac{4 + 2\lambda}{10}\right)^2\right) = 5 $$ $$ x^2 = \frac{5}{1 + \left(\frac{4 + 2\lambda}{10}\right)^2} $$ 13. **Calculate $F(x,y)$ at these points:** $$ F = 2x^2 + 10xy + 3y^2 $$ Substitute $y$: $$ F = 2x^2 + 10x \left(-\frac{4 + 2\lambda}{10} x\right) + 3 \left(-\frac{4 + 2\lambda}{10} x\right)^2 $$ $$ = 2x^2 - (4 + 2\lambda) x^2 + 3 \left(\frac{(4 + 2\lambda)^2}{100} x^2 \right) $$ $$ = x^2 \left(2 - (4 + 2\lambda) + \frac{3(4 + 2\lambda)^2}{100} \right) $$ 14. **Plug in $x^2$ from step 12 and simplify to find extrema values.** 15. **Classify extrema:** Since the problem is constrained on a circle, these values correspond to maxima or minima on the boundary. **Final answer:** The extrema occur at points on the circle $x^2 + y^2 = 5$ where $y = -\frac{4 + 2\lambda}{10} x$ with $\lambda = \frac{-5 \pm \sqrt{101}}{2}$, and the corresponding $F(x,y)$ values are given by the expression in step 13 evaluated at these points.