<s>[INST] Hi, I need help with the assignment, Find the point $P$ on the sphere $x^2 + y^2 + z^2 − 8 x − 4 y − 10 z = 4$ which is furthest from the plane Find the point $P$ on the sphere
$x^2 + y^2 + z^2 − 8x − 4y − 10z = 4$
which is furthest from the plane
$6x + 2y − 9z = −23$.
 [/INST] Let the point $P \equiv (p, q, r)$ 
The distance of $P$ from $6x+2y−9z=−23$ is given by $\frac{6p+2q−9r+23}{\sqrt{6^2+2^2+9^2}} = \frac{6p+2q−9r+23}{11}$
In other words, we are required to maximize $\frac{6p+2q−9r+23}{11}$ subject to $p^2+q^2+r^2−8p−4q−10r-4= 0$
This is equivalent to the following problem:
Maximize $6p+2q−9r$ subject to $p^2+q^2+r^2−8p−4q−10r-4= 0$
The solution is standard:
Let $f = (6p+2q−9r) + \lambda (p^2+q^2+r^2−8p−4q−10r-4)$
Set $\frac{\partial f}{\partial p} = \frac{\partial f}{\partial q} = \frac{\partial f}{\partial r} = \frac{\partial f}{\partial \lambda} = 0$ and solve.
You'll get $x = \frac{8\lambda -6}{2\lambda}$, $y = \frac{4\lambda -2}{2\lambda}$ and $z = \frac{10\lambda +9}{2\lambda}$ from the first three equations.
Substitute these values into the fourth equation (which is same as the equation of the sphere) and solve for $\lambda$. The solution is $\lambda = \pm \frac{11}{14}$
Corresponding to the two values of $\lambda$, the two possible values of $P$ are $(-\frac{42}{11}, -\frac{14}{11}, \frac{63}{11})$ and $(\frac{42}{11}, \frac{14}{11}, -\frac{63}{11})$ respectively.
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