Question

Use lagrange multipliers to find the point on the plane x â 2y + 3z = 6 that is closest to the point (0, 2, 4).

Answer 1

The distance between a point $(x,y,z)$ on the given plane and the point (0, 2, 4) is

$\sqrt{f(x,y,z)}=\sqrt{x^2+(y-2)^2+(z-4)^2}$

but since $\sqrt{f(x,y,z)}$ and $f(x,y,z)$ share critical points, we can instead consider the problem of optimizing $f(x,y,z)$ subject to $x-2y+3z=6$.

The Lagrangian is

$L(x,y,z,\lambda)=x^2+(y-2)^2+(z-4)^2+\lambda(x-2y+3z-6)$

with partial derivatives (set equal to 0)

$L_x=2x+\lambda=0\implies x=-\dfrac\lambda2$
$L_y=2(y-2)-2\lambda=0\implies y=2+\lambda$
$L_z=2(z-4)+3\lambda=0\implies z=4-\dfrac{3\lambda}2$
$L_\lambda=x-2y+3z-6=0\implies x-2y+3z=6$

Solve for $\lambda$:

$x-2y+3z=-\dfrac\lambda2-2(2+\lambda)+3\left(4-\dfrac{3\lambda}2\right)=6$
$\implies2=7\lambda\implies\lambda=\dfrac27$

which gives the critical point

$x=-\dfrac17,y=\dfrac{16}7,z=\dfrac{25}7$

We can confirm that this is a minimum by checking the Hessian matrix of $f(x,y,z)$:

$\mathbf H(x,y,z)=\begin{bmatrix}f_{xx}&f_{xy}&f_{xz}\\f_{yx}&f_{yy}&f_{yz}\\f_{zx}&f_{zy}&f_{zz}\end{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$

$\mathbf H$ is positive definite (we see its determinant and the determinants of its leading principal minors are positive), which indicates that there is a minimum at this critical point.

At this point, we get a distance from (0, 2, 4) of

$\sqrt{f\left(-\dfrac17,\dfrac{16}7,\dfrac{25}7\right)}=\sqrt{\dfrac27}$