<s>[INST] 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.
 [/INST] 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
$$-\sqrt{19} \le 10g - 3 \le \sqrt{19}$$
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