$ z = xy +\dfrac { 1 } { x } +\dfrac{1}{y}$

This article aims to find the**Â point on the surface**Â at which the **tangent plane is horizontal.**Â This article uses the**Â concept of the surface at which the tangent plane is horizontal. **To answer these questions, we must realize that the **horizontal plane is tangent to the curve**Â in space at **maximum, minimum, or saddle points.**Â Tangent planes to a surface are planes that touch the surface at a point and are **“parallel”**Â to the surface at a point.

**Expert Answer**

Determine **partial derivatives with respect**Â to $ x $ and $ y $ and set them equal to zero. Solve for $ x $ **partial with respect to**Â $ y $ and put the result back into partial with respect to $ y $ and put the result back into partial with respect to $ x $ to solve for $ y $, $ y $ can’t be zero because we can’t have a **zero denominator**Â in it, so $ y $ must be $ 1 $. Put $ 1 $ in the**Â equation for**Â $ y $ to find $ x $.

\[ z = x y + \dfrac { 1 } { x } + \dfrac { 1 } { y } \]

\[f_{ x } ( x , y ) Â = y – \dfrac { 1 } { x ^ { 2 } } = 0 \]

\[f_{ y } ( x , y ) = x – \dfrac { 1 } { y ^ { 2 } } Â = 0 \]

\[ x = \dfrac { 1 } { y ^ { 2 } } \]

\[ y – \dfrac { 1 } { \ dfrac { 1 } { y ^ { 2 } } } = Â 0 Â \]

\[-y^{2}+y = 0\]

\[y(-y+1)=0\]

\[y=1\]

\[x = \dfrac{1}{1^{2}}= 1\]

**Insert the point $(1,1)$ into $z$ and find the $3rd$ coordinate.**

\[ z(1,1) = 1.1 + \dfrac{1}{1}+\dfrac{1}{1} = 3\]

\[(x,y,z) = (1,1,3) \]

**Numerical Result**

The point on the surface at which the tangent plane is horizontal $ (x,y,z)=(1,1,3)$.

**Example**

**Find the point(s) on the surface at which the tangent plane is horizontal.**

**$ z = xy -\dfrac{1}{x} -\dfrac{1}{y}$**

**Solution**

Determine **partial derivatives with respect**Â to $ x $ and $ y $ and set them equal **to zero. **Solve for $ x $**partial with respect to $ y $ **and put the result back into**Â partial with respect to**Â $ y $ and put the result back into partial with respect to $ x $ to solve for $ y $, $ y $ can’t be**Â zero**Â because we can’t have a **zero denominator**Â in it, so $ y $ must be $ 1 $. Put $ 1 $ in the equation for $ x $ to find $ x $.

\[z = xy-\dfrac{1}{x}-\dfrac{1}{y} \]

\[f_{x}(x,y) = y+\dfrac{1}{x^{2}} = 0\]

\[f_{y}(x,y) = x+\dfrac{1}{y^{2}} = 0\]

\[x = -\dfrac{1}{y^{2}}\]

\[y+\dfrac{1}{\dfrac{1}{y^{2}}}= 0 \]

\[y^{2}+y = 0\]

\[y(y+1)=0\]

\[y=-1\]

\[x = -\dfrac{1}{-1^{2}}= -1\]

**Insert the point $(1,1)$ into $z$ and find the $3rd$ coordinate.**

\[ z(1,1) = (-1).(-1) – \dfrac{1}{-1}-\dfrac{1}{-1} = 3\]

\[(x,y,z) = (-1,-1,3) \]