$a+b+c=2,\ a+b^2+c^2=12$, what is the maximum value of c?

Question

Given that $a+b+c=2$ and $a+b^2+c^2=12$, what is the maximum value of c?

I have struggled with this problem for a few hours but couldn't solve.

How should I solve it?

Do you mean $a^2+b^2+c^2=12$? – In that case it is a duplicate of https://math.stackexchange.com/questions/669257/optimization-with-a-few-variables-amc-12-question. — Martin R, Nov 13 '17 at 12:23

score 2 · Accepted Answer · answered Nov 13 '17 at 12:31

Subtracting the first equation from the second, we get $b^2-b+c^2-c=10$, and that's equivalent with $(b-1/2)^2+(c-1/2)^2=21/2$. Since $(b-1/2)^2\ge0$, we have $(c-1/2)^2\le21/2$, i.e. $c\le1/2+\sqrt{21/2}$.

$a+b+c=2,\ a+b^2+c^2=12$, what is the maximum value of c?

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