This question aims to learn the basic methodology for **optimizing a mathematical function** (maximizing or minimizing).

**Critical points** are the points where the value of a function is either maximum or minimum. To calculate the **critical point(s)**, we equate the first derivative’s value to 0 and solve for the **independent variable**. We can use the** second derivative test **to find maxima/minima. For the **given question**, we can **minimize the distance function** **of the desired point** from the origin as explained in the below answer.

## Expert Answer

**Given:**

\[ y^{ 2 } \ = \ 9 \ + \ x \ z \]

**Let $ ( x, \ y, \ z ) $ be the point that is nearest to the origin. The distance of this point from the origin is calculated by:**

\[ d = \sqrt{ x^{ 2 } + y^{ 2 } + z^{ 2 } } \]

\[ \Rightarrow d^{ 2 } = x^{ 2 } + y^{ 2 } + z^{ 2 } \]

\[ \Rightarrow d^{ 2 } = x^{ 2 } + 9 + x z + z^{ 2 } \]

To find this point, **we simply need to minimize** this $ f(x, \ y, \ z) \ = \ d^{ 2 } $ function. **Calculating the first derivatives:**

\[ f_x = 2x + z \]

\[ f_z = x + 2z \]

Finding **critical points** by putting $ f_x $ and $ f_z $ equal to zero:

\[ 2x + z = 0\]

\[ x + 2z = 0\]

Solving the above system yields:

\[ x = 0\]

\[ z = 0\]

Consequently:

\[ y^{ 2 } = 9 + xz = 9 + (0)(0) = 0 \]

\[ \Rightarrow = y = \pm 3 \]

Hence, the** two possible critical points** are $ (0, 3, 0) $ and $ (0, -3, 0) $. **Finding the second derivatives**:

\[ f_{xx} = 2 \]

\[ f_{zz} = 2 \]

\[ f_{xz} = 1 \]

\[ f_{zx} = 1 \]

Since **all second derivatives are positive**, the calculated **critical points are at a minimum**.

## Numerical Result

Points Closest to the origin = $ (0, 0, 5)$ and $ (0, 0, -5) $

## Example

**Find the points on the surface $ z^2 = 25 + xy $ nearest to the origin.**

Here, the **distance function** becomes:

\[ d = \sqrt{ x^{ 2 } + y^{ 2 } + z^{ 2 } } \]

\[ \Rightarrow d^{ 2 } = x^{ 2 } + y^{ 2 } + z^{ 2 } \]

\[ \Rightarrow d^{ 2 } = x^{ 2 } + y^{ 2 } + 25 + xy \]

Calculating** first derivatives** and equating to zero:

\[ f_x = 2x + y \Rightarrow 2x + y = 0\]

\[ f_y = x + 2y \Rightarrow x + 2y = 0\]

Solving the above system yields:

\[ x = 0 \text{and} y = 0\]

Consequently:

\[ z^{ 2 } = 25 + xy = 25 \]

\[ \Rightarrow = z = \pm 5 \]

Hence, the **two possible critical points** are $ (0, 3, 0) $ and $ (0, -3, 0) $. **Finding the second derivatives**:

\[ f_{xx} = 2 \]

\[ f_{yy} = 2 \]

\[ f_{xy} = 1 \]

\[ f_{yx} = 1 \]

Since **all second derivatives are positive**, the calculated critical points are at a minimum.

**Points Closest to the origin = $ (0, 0, 5) $ and $ (0, 0, -5) $**