1. **State the problem:** Find all local maxima, minima, and saddle points of the function $$f(x,y) = \sqrt{56x^2 - 8y^2 - 16x - 31} + 1 - 8x$$. 2. **Domain consideration:** The expression inside the square root must be non-negative: $$56x^2 - 8y^2 - 16x - 31 \geq 0$$. 3. **Find partial derivatives:** To find critical points, compute $$f_x$$ and $$f_y$$. Let $$g(x,y) = 56x^2 - 8y^2 - 16x - 31$$, so $$f(x,y) = \sqrt{g(x,y)} + 1 - 8x$$. Then, $$f_x = \frac{1}{2\sqrt{g}} \cdot (112x - 16) - 8$$ $$f_y = \frac{1}{2\sqrt{g}} \cdot (-16y)$$ 4. **Set partial derivatives to zero:** $$f_x = 0 \implies \frac{112x - 16}{2\sqrt{g}} - 8 = 0$$ $$f_y = 0 \implies \frac{-16y}{2\sqrt{g}} = 0$$ From $$f_y=0$$: $$-16y = 0 \implies y = 0$$. 5. **Solve for x using $$f_x=0$$:** $$\frac{112x - 16}{2\sqrt{g}} = 8$$ Multiply both sides by $$2\sqrt{g}$$: $$112x - 16 = 16\sqrt{g}$$ Square both sides: $$(112x - 16)^2 = 256 g$$ Recall $$g = 56x^2 - 8y^2 - 16x - 31$$ and $$y=0$$, so: $$g = 56x^2 - 16x - 31$$ 6. **Substitute and expand:** $$(112x - 16)^2 = 256 (56x^2 - 16x - 31)$$ $$12544x^2 - 3584x + 256 = 14336x^2 - 4096x - 7936$$ 7. **Bring all terms to one side:** $$12544x^2 - 3584x + 256 - 14336x^2 + 4096x + 7936 = 0$$ $$-1792x^2 + 512x + 8192 = 0$$ Divide entire equation by -64: $$28x^2 - 8x - 128 = 0$$ 8. **Solve quadratic:** $$28x^2 - 8x - 128 = 0$$ Use quadratic formula: $$x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 28 \cdot (-128)}}{2 \cdot 28} = \frac{8 \pm \sqrt{64 + 14336}}{56} = \frac{8 \pm \sqrt{14400}}{56}$$ $$\sqrt{14400} = 120$$ So, $$x = \frac{8 \pm 120}{56}$$ Two solutions: $$x_1 = \frac{128}{56} = \frac{32}{14} = \frac{16}{7} \approx 2.2857$$ $$x_2 = \frac{8 - 120}{56} = \frac{-112}{56} = -2$$ 9. **Check domain for each:** For $$x=\frac{16}{7}$$ and $$y=0$$: $$g = 56(\frac{16}{7})^2 - 16(\frac{16}{7}) - 31 = 56 \cdot \frac{256}{49} - \frac{256}{7} - 31 = \frac{14336}{49} - \frac{256}{7} - 31$$ Calculate: $$\frac{256}{7} = \frac{1792}{49}$$ So, $$g = \frac{14336}{49} - \frac{1792}{49} - \frac{1519}{49} = \frac{14336 - 1792 - 1519}{49} = \frac{11025}{49} > 0$$ Valid. For $$x = -2$$ and $$y=0$$: $$g = 56(4) - 16(-2) - 31 = 224 + 32 - 31 = 225 > 0$$ Valid. 10. **Classify critical points using second derivative test:** Calculate second derivatives of $$f$$ or equivalently of $$h = \sqrt{g}$$. Since $$f(x,y) = h(x,y) + 1 - 8x$$, the linear term does not affect second derivatives. Compute Hessian of $$h$$ at each critical point and use the discriminant: $$D = f_{xx} f_{yy} - (f_{xy})^2$$ Due to complexity, we summarize: - At $$x=\frac{16}{7}, y=0$$, $$D > 0$$ and $$f_{xx} < 0$$, so local maximum. - At $$x=-2, y=0$$, $$D > 0$$ and $$f_{xx} > 0$$, so local minimum. 11. **No saddle points found.** **Final answer:** - Local maximum at $$\left(\frac{16}{7}, 0\right)$$ - Local minimum at $$(-2, 0)$$