The plane x+y+2z=18 intersects the paraboloid z=x^2+y^2 in an ellipse.Find the points on this ellipse that are nearest to and farthest from the origin

alesterp 2021-09-07 Answered
The plane \(\displaystyle{x}+{y}+{2}{z}={18}\) intersects the paraboloid \(\displaystyle{z}={x}^{{2}}+{y}^{{2}}\) in an ellipse. Find the points on this ellipse that are nearest to and farthest from the origin.

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Expert Answer

SoosteethicU
Answered 2021-09-08 Author has 19044 answers

It suffices to find the extreme values of the square of the distance to the origin \(\displaystyle{D}={x}^{{2}}+{y}^{{2}}+{z}^{{2}}\), subject to the constraints \(\displaystyle{g}={x}+{y}+{2}{z}={18}{\quad\text{and}\quad}{h}={x}^{{2}}+{y}^{{2}}-{z}={0}\). \(\mu\)

By Lagrange Multipliers,\(\triangledown D = \lambda \triangledown g + \mu \triangledown h.\) \(\displaystyle\Rightarrow\)\(<2x, 2y, 2z> = \lambda<1, 1, 2> + \mu<2x, 2y, -1>.\)

Equating like entries, \(\displaystyle{2}{x}=\lambda+{2}\mu{x}\)
\(\displaystyle{2}{y}=\lambda+{2}\mu{y}\)
\(\displaystyle{2}{z}={2}\lambda-\mu\)

Since the last equation can be rewritten as \(\displaystyle\mu={2}\lambda-{2}{z}\), we now have \(\displaystyle{2}{x}=\lambda+{2}{x}{\left({2}\lambda-{2}{z}\right)}=\Rightarrow{2}{x}=\lambda{\left({4}{x}+{1}\right)}-{4}{x}{z}\)
\(\displaystyle{2}{y}=\lambda+{2}{y}{\left({2}\lambda-{2}{z}\right)}=\Rightarrow{2}{y}=\lambda{\left({4}{y}+{1}\right)}-{4}{y}{z}\)
So,\(\displaystyle\frac{{{2}{x}+{4}{x}{z}}}{{{4}{x}+{1}}}=\lambda=\frac{{{2}{y}+{4}{y}{z}}}{{{4}{y}+{1}}}\Rightarrow{2}{x}\frac{{{1}+{2}{z}}}{{{4}{x}+{1}}}={2}{y}\frac{{{1}+{2}{z}}}{{{4}{y}+{1}}}.\) 

*If \(\displaystyle{1}+{2}{z}={0}\le\Rightarrow{z}=-\frac{{1}}{{2}},{t}{h}{e}{n}\ {h}={x}^{{2}}+{y}^{{2}}=-\frac{{1}}{{2}}\) has no solution.

So, we can ignore this case.

Now, we can freely divide both sides by \(\displaystyle{\left({1}+{2}{z}\right)}\):

\(\displaystyle\frac{{{2}{x}}}{{{4}{x}+{1}}}=\frac{{{2}{y}}}{{{4}{y}+{1}}}\)
\(\displaystyle\Rightarrow{2}{x}{\left({4}{y}+{1}\right)}={2}{y}{\left({4}{x}+{1}\right)}\)
\(\displaystyle\Rightarrow{x}={y}.\)

Substitute this into the constraint equations:
\(\displaystyle{x}+{x}+{2}{z}={18}\Rightarrow{x}+{z}={9}\)
\(\displaystyle{x}^{2}+{x}^{2}-{z}={0}\Rightarrow{z}={2}{x}^{2}\)

So,\(\displaystyle{x}+{2}{x}^{2}={9}.\)
\(\displaystyle\Rightarrow{2}{x}^{2}+{x}-{9}={0}\)
\(\displaystyle\Rightarrow{x}=\frac{{-{1}+\sqrt{{{73}}}}}{{4}}{\quad\text{or}\quad}\frac{{-{1}-\sqrt{{{73}}}}}{{4}}\)

So, \(\displaystyle{\left({x},{y}\right)}={\left(\frac{{-{1}+\sqrt{{{73}}}}}{{4}},\frac{{-{1}+\sqrt{{{73}}}}}{{4}}\right)}{\quad\text{or}\quad}{\left(\frac{{-{1}-\sqrt{{{73}}}}}{{4}},\frac{{-{1}-\sqrt{{{73}}}}}{{4}}\right)}\),

since \( y = x\)
\(\displaystyle\Rightarrow{\left({x},{y},{z}\right)}={\left(\frac{{-{1}+\sqrt{{{73}}}}}{{4}},\frac{{-{1}+\sqrt{{{73}}}}}{{4}},\frac{{{37}-\sqrt{{{73}}}}}{{4}}\right)}{\quad\text{or}\quad}{\left(\frac{{-{1}-\sqrt{{{73}}}}}{{4}},\frac{{-{1}-\sqrt{{{73}}}}}{{4}},\frac{{{37}+\sqrt{{{73}}}}}{{4}}\right)}\), using either constraint.
Finally, substitute this into D:
\(\displaystyle{D}{\left(\frac{{-{1}+\sqrt{{{73}}}}}{{4}},\frac{{-{1}+\sqrt{{{73}}}}}{{4}},\frac{{{37}-\sqrt{{{73}}}}}{{4}}\right)}=\frac{{{1590}-{78}\sqrt{{{3}}}}}{{16}}\) <---Closest
\(\displaystyle{D}{\left(\frac{{-{1}-\sqrt{{{73}}}}}{{4}},\frac{{-{1}-\sqrt{{{73}}}}}{{4}},\frac{{{37}+\sqrt{{{73}}}}}{{4}}\right)}=\frac{{{1590}+{78}\sqrt{{{3}}}}}{{16}}\) <---Farthest.

 

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