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Levart [38]
3 years ago
13

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

Mathematics
1 answer:
Inga [223]3 years ago
4 0
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}).
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