Yes, it is possible to find the furthest point from $(1, 0)$ on the ellipse $8x^2 + y^2 = 8$ using the Lagrange Multiplier method. The Lagrange Multiplier method is a powerful tool for finding the maximum or minimum values of a function subject to a constraint.

Steps to Solve the Problem

1. Define the Function and Constraint:

   • The function to maximize is the square of the distance from $(1, 0)$:
     $f(x, y) = (x - 1)^2 + y^2$
   • The constraint is the equation of the ellipse: $g(x, y) = 8x^2 + y^2 - 8 = 0$

2. Set Up the Lagrange Equations: The Lagrange Multiplier method requires solving: $\nabla f(x, y) = \lambda \nabla g(x, y)$ This means: $\frac{\partial f}{\partial x} = \lambda \frac{\partial g}{\partial x}, \quad \frac{\partial f}{\partial y} = \lambda \frac{\partial g}{\partial y}, \quad g(x, y) = 0$

3. Compute Gradients:

   • Gradient of $f(x, y)$:
     $\nabla f(x, y) = (2(x - 1), 2y)$
   • Gradient of $g(x, y)$:
     $\nabla g(x, y) = (16x, 2y)$

   Substituting into the Lagrange system: $2(x - 1) = \lambda (16x), \quad 2y = \lambda (2y), \quad 8x^2 + y^2 = 8$

4. Solve the System of Equations: From the equations:

   • From $2(x - 1) = \lambda (16x)$: $\lambda = \frac{2(x - 1)}{16x} = \frac{x - 1}{8x}, \quad \text{provided } x \neq 0$
   • From $2y = \lambda (2y)$: $\lambda = 1 \quad \text{or} \quad y = 0$
   • Substitute $\lambda = 1$ into $\lambda = \frac{x - 1}{8x}$: $1 = \frac{x - 1}{8x} \implies 8x = x - 1 \implies 7x = -1 \implies x = -\frac{1}{7}$
   • If $y = 0$, substitute into $8x^2 + y^2 = 8$: $8x^2 = 8 \implies x^2 = 1 \implies x = \pm 1$

5. Find the Points: The candidate points are: $(-1, 0), \quad (1, 0), \quad \left(-\frac{1}{7}, \pm \sqrt{8 - 8\left(-\frac{1}{7}\right)^2}\right)$

6. Determine the Furthest Point: Compute $f(x, y)$ for each candidate point to find the maximum value.

Would you like me to perform the detailed calculations for the furthest distance or clarify any specific step?

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Related Questions:

1. How does the Lagrange Multiplier method work for optimization problems?
2. What are the geometric implications of using gradients in optimization?
3. Can the Lagrange Multiplier method find both minima and maxima simultaneously?
4. How do we deal with equality constraints in optimization problems?
5. Why do we square the distance function rather than use the distance directly?

Tip: Always double-check the constraint values after solving to ensure the points lie on the given curve or surface.