This question **aims** to explain the concepts of **maxima** and **minima.** Formulas to **calculate** the **extreme** values of the **function.** Further, it explains how to calculate the **distance** between the points.

In mathematics, the **length** of the line segment between the two **points** is the Euclidean **distance** between two **points.** The **Pythagorean** theorem is used to calculate the **distance** from the **cartesian coordinates** of the point. It is also called the **Pythagorean** distance.

The **largest** and **smallest** value of the function is called its **maxima** and **minima** respectively either for the entire **domain** or the given **range.** They are also called the **extrema** of the function.

## Expert Answer

Let us assume the **point** $B(x,y, z)$ represents the **point** on the **cone.**

Finding the **distance** between the point $A(2,2, 0)$ and the point $B(x,y,z)$:

Inserting the values in the **distance** formula:

\[ d= \sqrt{ (x_2- x_1)^2+ (y_2- y_1)^2+ (z_2- z_1)^2} \]

\[d= \sqrt{ (x-2)^2+ (y-2)^2+ (z-0)^2} \]

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

**Inserting** the $z^2 = x^2 + y^2$ in the above equation:

\[d= \sqrt{ (x-2)^2+ (y-2)^2+ x^2 + y^2} \]

**Squaring** both sides:

\[d^2 =Â (x-2)^2+ (y-2)^2+ x^2 + y^2 \]

If we **minimize** $d^2$, we **minimize** the distance $d$ between the points $A(2,2, 0)$ and the point $B(x,y,z)$.

\[f’ = 0\]

\[ \dfrac{df}{dx} = \dfrac{d}{dx} (x-2)^2+ (y-2)^2+ x^2 + y^2 \]

\[ \dfrac{df}{dx} = 2(x-2)+ 2x \]

Putting $\dfrac{df}{dx}$ equals to $0$ and **solving** for $x$:

\[ 2x – 4 + 2x =0 \]

\[ 4x =4 \]

\[ x =1\]

**Similarly** solving for $y$:

\[ \dfrac{df}{dy} = \dfrac{d}{dy} (x-2)^2+ (y-2)^2+ x^2 + y^2 \]

\[ \dfrac{df}{dy} = 2(y-2)+ 2y \]

Putting $\dfrac{df}{dy}$ equals to $0$ and **solving** for $y$:

\[ 2y – 4 + 2y =0 \]

\[4y=4 \]

\[ y =1\]

Now **solving** $z^2 = x^2 + y^2$ by inserting the above **calculated** values of $x$ and $y$.

\[ z^2=1+1\]

\[ z^2=2\]

\[ z = \pm \sqrt{2} \]

## Numerical Results

The points on the cone $z^2= x^2 + y^2$ that are **closest** to the point $(2,2, 0)$ are $(1, 1, \sqrt{2})$ and $(1, 1, -\sqrt{2})$.

## Example

Find the **points** that are **closest** to the point $(4,2,0)$ on the **cone** $z^2 = x^2 + y^2$.

Assume the **point** $B(x,y z)$ to be the **point** on the **cone.**

The **distance** between the point $A(4,2, 0)$ and the **point** $B(x,y,z)$ is:

\[d= \sqrt{ (x-4)^2+ (y-2)^2+ (z-0)^2} \]

\[d= \sqrt{ (x-4)^2+ (y-2)^2+ z^2} \]

Inserting $z^2$:

\[d= \sqrt{ (x-4)^2+ (y-2)^2+ x^2 + y^2} \]

\[d^2 =Â (x-2)^2+ (y-2)^2+ x^2 + y^2 \]

**Minimizing** the **distance** $d$:

\[f’ =0\]

\[ \dfrac{df}{dx}= \dfrac{d}{dx} (x-4)^2+ (y-2)^2+ x^2 + y^2 =0 \]

\[ \dfrac{df}{dx}= 2(x-4)+ 2x =0\]

\[2x-8+2x=0\]

\[4x =8\]

\[ x =2\]

**Similarly** solving for $y$:

\[\dfrac{df}{dy}= \dfrac{d}{dy} (x-4)^2+ (y-2)^2+ x^2 + y^2 =0 \]

\[\dfrac{df}{dy}=2(y-2)+ 2y=0 \]

\[2y-4+2y=0\]

\[ 4y=4\]

\[ y =1\]

Now **solving** $z^2 = x^2 + y^2$ by **inserting** the above **calculated** values of $x$ and $y$.

\[z^2=2^2 +1\]

\[z^2=5\]

\[z= \pm \sqrt{5}\]

**Closest** points are $(2,1, \sqrt{5})$ and $(2,1, -\sqrt{5})$