Find the point on the surface $x^2+2y^2-z^2-1$ that is nearest to the origin.
I know we have to use Lagrange's method of multipliers here. But I am having trouble finding a specific point in this case.
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2$\begingroup$ You want $x^2+2y^2-z^2-1=0$ to define a surface. $\endgroup$– JulienMar 18, 2013 at 11:47
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2 Answers
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You want to minimize $x^2+y^2+z^2$ assuming $z^2=x^2+2y^2+1$. So you have $x^2+y^2+z^2=2x^2+3y^2+1$. It is obvious that the minimal value of $2x^2+3y^2+1$ is attained for $x=y=0$.
answered Mar 18, 2013 at 14:25
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We want to minimise the function $$ \left\|\cdot\right\|:(x,y,z)\mapsto\left\|(x,y,z)\right\|=\sqrt{x^2+y^2+z^2} $$ subject to the constraint $$ x^2+2y^2-z^2-1=0. $$
By the method of Lagrange multipliers this task amounts to minimizing the function $$ (x,y,z,\lambda)\mapsto\sqrt{x^2+y^2+z^2}-\lambda(x^2+2y^2-z^2-1). $$ Setting the four partial derivatives equal to zero results in a system of four equations with four unknowns, which has four real solutions $(x,y,z,\lambda)$. These are $$ \left\{(-1,0,0,1/2),(1,0,0,1/2),(0,-1/\sqrt{2},0,1/(2\sqrt{2})),(0,1/\sqrt{2},0,1/(2\sqrt{2}))\right\}. $$
The points on the surface nearest to the origin are thus $(0,\pm1/\sqrt{2},0)$.
You may want to check that the critical points so computed are indeed local minima rather than, say, maxima.
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6$\begingroup$ would be easier minimizing $||\cdot||^2$ rather than $||\cdot||$ $\endgroup$– yohBSMar 18, 2013 at 11:14
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$\begingroup$ No, I don't think squaring would be easier than this. $\endgroup$ Apr 2, 2013 at 14:45
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$\begingroup$ sorry to butt in, but what is this symbol ∥⋅∥ should be read as? thank u. @yohBS $\endgroup$ Oct 1, 2014 at 16:35
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$\begingroup$ @alethiologist norm en.wikipedia.org/wiki/Norm_(mathematics)#Notation $\endgroup$– yohBSOct 7, 2014 at 1:41