Minimize g(x, y)=max{12-x,8-y} over x+y=10

Question

Let $g (x, y)=\max\{12-x,8-y\}$. Find the minimum value of $g(x,y)$ as $(x, y)$ varies over the line $x+y=10$.

1.5

2.1

3.7

4.3

My attempt

Without use of Lagrange method of multiplier , I see graphically that at $(7,3)$ $12-x$ and $8-y$ both equal $5$. And $5$ seems to be minimum value graphically.

Answer 1

On $x+y=10$ we have $y=10-x$, so we have to minimize $\max \{12-x,x-2\}$. For $x \neq 7$, we have either $12-x>5$ or $x-2>5$. So $5$ is indeed the minimal value, obtained for $x=7$.