# Find the points on the cone z^2=x^2+y^2 that are closest to the point (4,2

floymdiT 2021-10-18 Answered
Find the points on the cone ${z}^{2}={x}^{2}+{y}^{2}$ that are closest to the point (4,2,0)
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## Expert Answer

tabuordy
Answered 2021-10-19 Author has 91 answers

Using the distance formula we can say that the distance between a general point (x,y,z) and (4,2,0) is given by
$D=\sqrt{{\left(x-4\right)}^{2}+{\left(y-2\right)}^{2}+{\left(z-0\right)}^{2}}$
Since the points are on the cone we can replace ${z}^{2}$ with ${x}^{2}+{y}^{2}$
$D=\sqrt{{\left(x-4\right)}^{2}+{\left(y-2\right)}^{2}+{x}^{2}+{y}^{2}}$
Consider the following function
$f\left(x,y\right)={\left(x-4\right)}^{2}+{\left(y-2\right)}^{2}+{x}^{2}+{y}^{2}$
Note that D(x,z) will be minimum when f(x,z) is minimum, so we will work with f(x,z) for conveince
To find the minimum value of f(x,y), we will us the second derivative test.
Second derivatives test:
If the second partial derivatives of f are continuous on a disk containing the critical point (a,b) , and
$D\left(a,b\right)={f}_{xx}\left(a,b\right){f}_{yy}\left(a,b\right)-{\left[{f}_{xy}\left(a,b\right)\right]}^{2}$
Then we can conclude that:
1) $D>0$ and ${f}_{xx}\left(a,b\right)<0⇒$ a local minimum
2) $D>0$ and ${f}_{xx}\left(a,b\right)<0⇒$ a local maximum
3) $D<0⇒$ no local maximuum/minimum
We can find the local extrema using the Second Derivatives Tes from the section. Note that
$f\left(x,y\right)={\left(x-4\right)}^{2}+{\left(y-2\right)}^{2}+{x}^{2}+{y}^{2}$
is a polynomial function and hence is continuous everywhere in its domain. Thus the theorem applies.
Firstly, we will need the first and second partial derivatives of f:

Now we find the critical points of f. These are the points where both first partial derivatives are 0 at the same time. So let's solve the system of equations
$\left\{\begin{array}{l}2\left(x-4\right)+2x=0\\ 2\left(y-2\right)+2y=0\end{array}$
Solve the first equation
$\left(x-4\right)+x=0$
$2x=4$
Divide both sides by 2
$x=2$
Solve the first equation
$2\left(y-2\right)+2y=0$
$y-2+y=0$
$2y=2$
Divide both sides by 2
$y=1$
Therefore, the critical point is (2,1)
Now we just neet the value

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