Answer
The point $\left( { - 2\sqrt {\frac{3}{{35}}} , - \sqrt {\frac{3}{{35}}} ,2\sqrt {\frac{{15}}{7}} } \right)$ lies on the intersection of the plane $x + \frac{1}{2}y + \frac{1}{4}z = 0$ and the sphere ${x^2} + {y^2} + {z^2} = 9$ has the largest $z$-coordinate $2\sqrt {\frac{{15}}{7}} $.
Work Step by Step
Since we are finding the largest $z$-coordinate, our task is to maximize the function $f\left( {x,y,z} \right) = z$ subject to the two constraints $g\left( {x,y,z} \right) = x + \frac{1}{2}y + \frac{1}{4}z = 0$ and $h\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2} - 9 = 0$.
The Lagrange condition is
$\nabla f = \lambda \nabla g + \mu \nabla h$
(1) ${\ \ \ \ }$ $\left( {0,0,1} \right) = \lambda \left( {1,\frac{1}{2},\frac{1}{4}} \right) + \mu \left( {2x,2y,2z} \right)$
From the Lagrange condition we obtain three equations:
$0 = \lambda + 2\mu x$, ${\ \ \ }$ $0 = \frac{1}{2}\lambda + 2\mu y$, ${\ \ \ }$ $1 = \frac{1}{4}\lambda + 2\mu z$
From these equations we obtain
$\lambda = - 2\mu x = - 4\mu y = 4 - 8\mu z$
(2) ${\ \ \ \ }$ $ - 2\mu x = - 4\mu y$, ${\ \ \ }$ $ - 2\mu x = 4 - 8\mu z$
From equation (1) we notice that $\mu \ne 0$, otherwise it has no solution. Thus, the first equation of (2) gives $y = \frac{1}{2}x$.
Substituting $y = \frac{1}{2}x$ in $g$ gives
$x + \frac{1}{4}x + \frac{1}{4}z = 0$
$z = - 5x$
Substituting $y = \frac{1}{2}x$ and $z = - 5x$ in $h$ gives
${x^2} + \frac{1}{4}{x^2} + 25{x^2} - 9 = 0$
$\frac{{105{x^2}}}{4} = 9$
So, $x = \pm 2\sqrt {\frac{3}{{35}}} $.
Using $y = \frac{1}{2}x$ and $z = - 5x$, we obtain the critical points: $\left( {2\sqrt {\frac{3}{{35}}} ,\sqrt {\frac{3}{{35}}} , - 2\sqrt {\frac{{15}}{7}} } \right)$ and $\left( { - 2\sqrt {\frac{3}{{35}}} , - \sqrt {\frac{3}{{35}}} ,2\sqrt {\frac{{15}}{7}} } \right)$.
The extreme values corresponds to the two critical points are
$f\left( {2\sqrt {\frac{3}{{35}}} ,\sqrt {\frac{3}{{35}}} , - 2\sqrt {\frac{{15}}{7}} } \right) = - 2\sqrt {\frac{{15}}{7}} $
$f\left( { - 2\sqrt {\frac{3}{{35}}} , - \sqrt {\frac{3}{{35}}} ,2\sqrt {\frac{{15}}{7}} } \right) = 2\sqrt {\frac{{15}}{7}} $
From these results we conclude that the maximum of $f$ subject to the two constraints is $2\sqrt {\frac{{15}}{7}} $.
So, the point $\left( { - 2\sqrt {\frac{3}{{35}}} , - \sqrt {\frac{3}{{35}}} ,2\sqrt {\frac{{15}}{7}} } \right)$ lies on the intersection of the plane $x + \frac{1}{2}y + \frac{1}{4}z = 0$ and the sphere ${x^2} + {y^2} + {z^2} = 9$ has the largest $z$-coordinate $2\sqrt {\frac{{15}}{7}} $.
