To solve Question 5 from the uploaded document, here's how we can approach the problem:

Question 5 Analysis:

We need to find the point on the surface $z^2 = xy + 1$ that is nearest to the origin, using Lagrange's Method.

Step-by-Step Solution:

1. Objective Function:
   The distance of a point $(x, y, z)$ from the origin is given by:
   $D = \sqrt{x^2 + y^2 + z^2}.$
   To simplify the computation, we minimize $D^2 = x^2 + y^2 + z^2$ instead of $D$.

2. Constraint:
   The given constraint is: $z^2 = xy + 1.$

3. Form the Lagrange Function:
   Using Lagrange's multiplier method, let the function to minimize be: $F(x, y, z, \lambda) = x^2 + y^2 + z^2 + \lambda (xy + 1 - z^2),$ where $\lambda$ is the Lagrange multiplier.

4. Partial Derivatives:
   Find the partial derivatives of $F$ and set them equal to zero:

   • With respect to $x$: \frac{\partial F}{\partial x} = 2x + \lambda y = 0. \tag{1}
   • With respect to $y$: \frac{\partial F}{\partial y} = 2y + \lambda x = 0. \tag{2}
   • With respect to $z$: \frac{\partial F}{\partial z} = 2z - 2\lambda z = 0. \tag{3}
   • With respect to $\lambda$ (the constraint equation): xy + 1 - z^2 = 0. \tag{4}

5. Solve the System of Equations:

   • From $(3)$, $2z(1 - \lambda) = 0$:
     Either $z = 0$ or $\lambda = 1$.

   • Case 1: $z = 0$:
     Substitute $z = 0$ into $(4)$:
     $xy + 1 - 0^2 = 0 \implies xy = -1.$
     Substituting $z = 0$ into $(1)$ and $(2)$, we get:
     $\frac{\partial F}{\partial x} = 2x + \lambda y = 0 \quad \text{and} \quad \frac{\partial F}{\partial y} = 2y + \lambda x = 0.$
     Solving these gives $x = 0$ or $y = 0$, but these do not satisfy $xy = -1$. Hence, discard $z = 0$.

   • Case 2: $\lambda = 1$:
     Substitute $\lambda = 1$ into $(1)$ and $(2)$:
     2x + y = 0 \implies y = -2x, \tag{5}
     2y + x = 0 \implies x = -2y. \tag{6}
     From $(5)$ and $(6)$, solve for $x, y$:
     Substituting $y = -2x$ into $x = -2y$:
     $x = -2(-2x) \implies x = 4x \implies x = 0.$
     Thus, $x = 0$, $y = 0$.

     Substitute $x = 0, y = 0$ into $(4)$:
     $xy + 1 - z^2 = 0 \implies 0 + 1 - z^2 = 0 \implies z^2 = 1 \implies z = \pm 1.$

6. Conclusion:
   The points nearest to the origin are:
   $(x, y, z) = (0, 0, 1) \quad \text{and} \quad (x, y, z) = (0, 0, -1).$

Final Answer:

The points on the surface $z^2 = xy + 1$ nearest to the origin are:
$(0, 0, 1) \quad \text{and} \quad (0, 0, -1).$

Would you like a detailed explanation of Lagrange's Method or additional examples?

Related Questions:

1. What is the geometric interpretation of Lagrange's multiplier method?
2. Can the constraint $z^2 = xy + 1$ be represented in cylindrical coordinates?
3. How does minimizing $D^2$ simplify the problem over $D$?
4. What are the necessary conditions for maxima/minima in multivariable optimization?
5. How do we handle multiple constraints in Lagrange's method?

Tip:

When solving optimization problems, always check for edge cases (like $z = 0$ in this problem) to ensure a comprehensive solution.