Find maximum and minimum value of function $f(x,y,z)=y+z$ on the circle Find maximum and minimum value of function $$f(x,y,z)=y+z$$ on the circle $$x^2+y^2+z^2 = 1,3x+y=3$$
We have that $$y=3-3x$$ So we would like find minimum and maximum value of function $$g(x,z)=3-3x+z$$ on $$x^2+ (3(1-x))^2 + z^2 = 1$$ And then I was using Lagrange multipliers. I received that $$z=\frac{9-10x}{3}$$ and next I substituted it to circle equal. But the numbers are awful. 
This task comes from an exam so I suppose that is the easier way to solve it.
 A: Let's look at what happens when we perturb $(x,y,z)$ infinitesimally in the direction $(\delta x,\delta y,\delta z)$.
To stay on $x^2+y^2+z^2=1$, we need $x\,\delta x+y\,\delta y+z\,\delta z=0$ and to stay on $3x+y=3$, we need $3\delta x+\delta y=0$.
That is, we want to consider only $(\delta x,\delta y,\delta z)$ which are perpendicular to both $(x,y,z)$ and $(3,1,0)$.
We want to find the point at which $\delta f(x,y,z)=(0,1,1)\cdot(\delta x,\delta y,\delta z)=0$ for all $(\delta x,\delta y,\delta z)$ which are perpendicular to both $(x,y,z)$ and $(3,1,0)$. For that to be true, $(0,1,1)$ must be a linear combination of $(3,1,0)$ and $(x,y,z)$. This means that $(x,y,z)$ is also a linear combination of $(3,1,0)$ and $(0,1,1)$; that is, on the plane $x-3y+3z=0$.
Thus, we are looking for a point on the unit sphere that is on the planes $x-3y+3z=0$ and $3x+y=3$. The intersection of these two planes is the line $(0,3,3)+(-3,9,10)t$. So we need to solve
$$
\begin{align}
1&=(0-3t)^2+(3+9t)^2+(3+10t)^2\\
&=18+114t+190t^2
\end{align}
$$
that is $t=\dfrac{-57\pm\sqrt{19}}{190}$. Note that
$$
\begin{align}
f((0,3,3)+(-3,9,10)t)
&=(0,1,1)\cdot((0,3,3)+(-3,9,10)t)\\
&=6+19t
\end{align}
$$
Plugging in the values of $t$ give us the extreme values: $\dfrac{3+\sqrt{19}}{10}$ and $\dfrac{3-\sqrt{19}}{10}$
A: Another way.  Given that you have $g = 3- 3x + z$ to find extrema, with the constraint 
$1 = x^2 + 9(1-x)^2 + z^2 = (10x^2 -18x + 9) + z^2 = 10(x-\frac{9}{10})^2 + z^2 + \frac{9}{10}$
Thus the constraint is $100(x-\frac{9}{10})^2 + 10z^2 = 1$ 
Now we can let $x - \frac{9}{10} = \frac{\sin(t)}{10}$ and $z = \frac{\cos(t)}{\sqrt{10}}$ so that the constraint is satisfied.  
Then our objective becomes $g = 3 - 3(\frac{9}{10}+\frac{\sin(t)}{10})+ \frac{\cos(t)}{\sqrt{10}}$.  
$$\implies 10g-3 = - 3 \sin t + \sqrt{10} \cos t$$ 
Finding the extrema of this RHS should be easy using trigonometry or calculus, extrema of $a \sin t + b \cos t$ is $\pm \sqrt{a^2 + b^2}$.  So we have
$$-\sqrt{19} \le 10g - 3 \le \sqrt{19}$$
A: Using Lagrange/Lagrangian multiplier, we have 
$\displaystyle f(x,y,z)=y+z+a\{x^2+y^2+z^2-1\}$
$\displaystyle=3-3x+z+a\{x^2+(3-3x)^2+z^2-1\}=10ax^2-x(18a+3)+8a+az^2+3+8a$
$f_x=20ax-(18a+3)$
$f_z=1+2az$
$f_{xx}=20a,f_{zz}=2a,f_{xz}=f_{zx}=0$
Can you take it from here?
