Question

Find the shortest distance from the origin to the surface $\mathit{z}\mathbf{=}\mathit{x}\mathit{y}\mathbf{+}\mathbf{5}$.

Short answer

Hence, the shortest distance is 3 units.

Step-by-step solution

Distance Formula

The formula to find out the distance of any point $\left(x,y,z\right)$in a three-dimensional space from the origin $\left(0,0,0\right)$is given by:

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

Step 2:Find the function

The given surface is:$z=xy+5$

Now, using the distance formula, we have:

role="math" localid="1664343640667" $\begin{array}{rcl}d& =& \sqrt{x^2+y^2+z^2}\\ d^2& =& x^2+y^2+z^2=f .................\left(\text{say}\right)\end{array}$

Put $z=xy+5$in this formula as follow:

$\begin{array}{rcl}f& =& x^2+y^2+{\left(xy+5\right)}^2\\ & =& x^2+y^2+x^2{y}^2+10xy+25\end{array}$

Find the Partial Derivative

Find the differential and equate it to zero as follows:

$\begin{array}{rcl}\frac{\partial f}{\partial x}& =& 0\\ 2x+2\left(xy+5\right)y& =& 0\\ x+x{y}^2+5y& =& 0\end{array}$

Similarly, we get:

$\begin{array}{rcl}\frac{\partial f}{\partial y}& =& 0\\ 2y+2\left(xy+5\right)x& =& 0\\ y+x^{2}y+5x& =& 0\end{array}$

Solve equations

Now, solve both the equations as follows:

$\begin{array}{rcl}\left(x-y\right)+\left(x{y}^2-x^{2}y\right)+5\left(y-x\right)& =& 0\\ -\left(y-x\right)+xy\left(y-x\right)+5\left(y-x\right)& =& 0\\ xy\left(y-x\right)+4\left(y-x\right)& =& 0\\ \left(xy+4\right)\left(y-x\right)& =& 0\end{array}$

Solving further, we get:

$x=y\text{or}xy=-4$

At $x=y$:

$\begin{array}{rcl}x+x^3+5x& =& 0\\ x\left(x^2+6\right)& =& 0\\ x& =& 0 ..........\left\{x^2+6\ne 0\right\}\end{array}$

And at $y=-\frac{4}{x}$:

$\begin{array}{rcl}x+x{\left(-\frac{4}{x}\right)}^2+5\left(-\frac{4}{x}\right)& =& 0\\ \left(x-\frac{4}{x}\right)& =& 0\\ x^2& =& 4\\ x& =& \pm 2 ..........\left\{x^2+6\ne 0\right\}\end{array}$

Step 5:Find Distance

Here, the three solutions are $\left(0,0\right),\left(2,-2\right)\text{and}\left(-2,2\right)$. Put these points in the equation $z=xy+5$, and we get:

$$\begin{gathered} {\overline{)z}}_{(0,0)}=0·0+5=5 \\ {\overline{)z}}_{(2,-2)}=2·\left(-2\right)+5=1 \\ {\overline{)z}}_{(-2,2)}=\left(-2\right)·2+5=1 \end{gathered}$$

Now, the points are: $\left(0,0,5\right),\left(2,-2,1\right)\text{and}\left(-2,2,1\right)$

The distance of these points from the origin will be:

$$\begin{gathered} {\overline{)d}}_{(0,0,5)}=\sqrt{0^2+0^2+5^2}=5 \\ {d}_{(2,-2,1)}=\sqrt{2^2+{\left(-2\right)}^2+1^2}=3 \\ {d}_{(-2,2,1)}=\sqrt{{\left(-2\right)}^2+2^2+1^2}=3 \end{gathered}$$

Hence, the shortest distance is 3 units.