extrema with constraints (lagrange?)

Question

I'd like to find the point of $E: 2x+3y+z = 14$ which has the smallest distance to the point of origin (0,0,0).

I think I have $ d(x,y,z) = \sqrt{x^2+y^2+z^2}$ with constraint $2x+3y+z = 14$.

What exactly do I have to do in such a case? Is the lagrange multiplier the only way?

Thank you very much for any kind of help! =)

Answer 1

Hint

Let $d^2(x,y,z)=f(x,y,z)=x^2+y^2+z^2$ and $g(x,y,z)=2x+3y+z-14=0$ and use Lagrange multipliers as follow

$$\begin{cases} f_x=\lambda g_x \\ f_y=\lambda g_y\\ f_z=\lambda g_z\\ g(x,y,z)=0 \end{cases} $$ After some algebra I got $$x=2, y=3, z=1$$

Answer 2

Then Let g(x,y,z)=2x+3y+z-14=0 find$$f_{x}=L*g_{x}$$ $$f_{y}=L*g_{y}$$ $$f_{z}=L*g_{z}$$ Where $f(x,y,z)=x^{2}+y^{2}+z^{2}$ You have four simultaneous equation with four unknowns and you can thus solve the equation.

You also have another method

$1)$Replace the constraint equation in the equation to be maximized or minimized

$2)$Then you have a new function$F(x,y)$

$3)$Find the partial derivatives and set to 0

$4)$Solve the simultaneous equations to get $x$ & $y$

I didn't use notation "lamda" because I couldn't render the symbol. I used L instead

Answer 3

If you want to do it with Lagrange multipliers, it's better to consider $d^2$, since $d$ is always positive.

But that's not the only way in this case: because the constraint is that the point lie in a plane, you can recall that the shortest distance vector will be perpendicular to the plane; but we know what this is: a plane is given in the form $$ {\bf n} \cdot {\bf x} = a, $$ where ${\bf n}$ is perpendicular to the plane, though not necessarily of unit length. Hence all we have to do is solve $$ {\bf n} \cdot {\bf x} = a \quad {\bf x}=t{\bf n} $$ simultaneously: this is the intersection of the plane and the line perpendicular to it starting at $0$. The length of the resulting vector is then the shortest distance. So $$ t \lVert n \rVert^2 = a, $$ and the vector is $ {\bf x} = \frac{a}{\lVert n \rVert^2}{\bf n} $, and the distance is $ \lvert a \rvert / \lVert {\bf n} \rVert $.

Putting the numbers in, $$ \frac{\lvert a \rvert}{\lVert {\bf n} \rVert} = \frac{14}{\sqrt{2^2+3^2+1^2}} = \frac{14}{\sqrt{14}} = \sqrt{14}, $$ (which thankfully does agree with the answer from doing Lagrange multipliers, so I believe it).

Answer 4

Hint

There is another way which could be the minimization of the square of the distance and the elimination of one variable from the equality constraint. So, eliminate $z$ from the constraint $$z=14-2x-3y$$ and plug it into $x^2+y^2+z^2$. So, you want to minimize $$F=x^2+y^2+(14-2x-3y)^2$$ Compute the partials $$F'_x=10 x+12 y-56$$ $$F'_y=12 x+20 y-84$$ Say that at the extremum the partial derivatives are zero.