# Maximum and minimum of a three-variable system

Let $a,b,c$ be reals such that $a-b+c=3, a^2+b^2+c^2=4$. Find the maximum and minimum of $\sqrt{2}a + \sqrt{2}b + 3c$.

Don't use coordinates or Lagrange multipliers in order to solve this. Are there any algebraic ways to solve this that don't involve Calculus?

If Lagrange Multipliers is banned, what we can do is set $\sqrt{2}a+\sqrt{2}b+3c=k$, and with $a-b+c=3$, we can write $$a=\frac{k+3\sqrt{2}-(3+\sqrt{2})c}{2\sqrt{2}}, b=-3+\frac{k+3\sqrt{2}-(3+\sqrt{2})c}{2\sqrt{2}}+c$$
Plug these values into $a^2+b^2+c^2=4$. You will have a quadratic (very dirty) in $c,k$.
Take the determinant to find the range of $k$.
Solving gives the minimum as $-0.1622$ and maximum as $6.1622$.