Question

Use lagrange multipliers to find the points on the cone z2 = x2 + y2 that are closest to (16, 6, 0).

Answer 1

Minimize $(x-16)^2+(y-6)^2+z^2$ subject to $z^2=x^2+y^2$. The Lagrangian would be

$L(x,y,z,\lambda)=(x-16)^2+(y-6)+z^2+\lambda(z^2-x^2-y^2)$

and has partial derivatives

$\begin{cases}L_x=2(x-16)-2\lambda x\\L_y=2(y-6)-2\lambda y\\L_z=2z+2\lambda z\\L_\lambda z^2-x^2-y^2\end{cases}$

Setting each partial derivative to 0, we have

$\begin{cases}2(x-16)-2\lambda x=0\implies (1-\lambda)x=16\\2(y-6)-2\lambda y=0\implies(1-\lambda)y=6\\2z+2\lambda z=0\implies(1+\lambda)z=0\\z^2-x^2-y^2=0\implies z^2=x^2+y^2\end{cases}$

From the third equation, it follows that either $\lambda=-1$ or $z=0$. In the second case, we arrive at a contradiction:

$z^2=x^2+y^2=0\implies x=y=0$

since both $x^2$ and $y^2$ must be non-negative, yet this would mean e.g. $(1-\lambda)x=0=16$. So it must be that $\lambda=-1$.

The first and second equations then tell us that

$(1-\lambda)x=2x=16\implies x=8$
$(1-\lambda)y=2y=6\implies y=3$

from which we obtain $z^2=8^2+3^2=73\implies z=\pm\sqrt{73}$.

Thus the points on the cone closest to (16, 6, 0) are $(8,3,\pm\sqrt{73})$.