1. **State the problem:** Find the absolute extrema of the function $$f(x,y,z) = 2x - 3y + z - 1$$ subject to the constraint $$x^2 + y^2 + z^2 = 14$$. 2. **Method:** Use Lagrange multipliers to find extrema of $$f$$ under the constraint. 3. **Set up Lagrange function:** $$L(x,y,z,\lambda) = 2x - 3y + z - 1 - \lambda (x^2 + y^2 + z^2 - 14)$$ 4. **Find partial derivatives and set to zero:** $$\frac{\partial L}{\partial x} = 2 - 2\lambda x = 0 \implies 2 = 2\lambda x \implies x = \frac{1}{\lambda}$$ $$\frac{\partial L}{\partial y} = -3 - 2\lambda y = 0 \implies -3 = 2\lambda y \implies y = -\frac{3}{2\lambda}$$ $$\frac{\partial L}{\partial z} = 1 - 2\lambda z = 0 \implies 1 = 2\lambda z \implies z = \frac{1}{2\lambda}$$ $$\frac{\partial L}{\partial \lambda} = -(x^2 + y^2 + z^2 - 14) = 0 \implies x^2 + y^2 + z^2 = 14$$ 5. **Substitute expressions for $$x,y,z$$ into constraint:** $$\left(\frac{1}{\lambda}\right)^2 + \left(-\frac{3}{2\lambda}\right)^2 + \left(\frac{1}{2\lambda}\right)^2 = 14$$ Simplify: $$\frac{1}{\lambda^2} + \frac{9}{4\lambda^2} + \frac{1}{4\lambda^2} = 14$$ $$\frac{1 + 9/4 + 1/4}{\lambda^2} = 14$$ $$\frac{1 + 2.5}{\lambda^2} = 14 \implies \frac{3.5}{\lambda^2} = 14$$ $$\lambda^2 = \frac{3.5}{14} = \frac{1}{4}$$ $$\lambda = \pm \frac{1}{2}$$ 6. **Find corresponding points:** - For $$\lambda = \frac{1}{2}$$: $$x = \frac{1}{1/2} = 2, \quad y = -\frac{3}{2 \times 1/2} = -3, \quad z = \frac{1}{2 \times 1/2} = 1$$ - For $$\lambda = -\frac{1}{2}$$: $$x = \frac{1}{-1/2} = -2, \quad y = -\frac{3}{2 \times -1/2} = 3, \quad z = \frac{1}{2 \times -1/2} = -1$$ 7. **Evaluate $$f$$ at these points:** - At $$(2, -3, 1)$$: $$f = 2(2) - 3(-3) + 1 - 1 = 4 + 9 + 1 - 1 = 13$$ - At $$(-2, 3, -1)$$: $$f = 2(-2) - 3(3) + (-1) - 1 = -4 - 9 - 1 - 1 = -15$$ 8. **Conclusion:** The absolute maximum of $$f$$ on the sphere is $$13$$ at $$(2, -3, 1)$$. The absolute minimum of $$f$$ on the sphere is $$-15$$ at $$(-2, 3, -1)$$.