Maximum under linear constraints

Statement

Find the maximum value of $x + y$ given that the constraints $2x + y \le 5$ and $x + 3y \le 10$ both hold.

Solution

We want $x+y$ as large as possible. It pays to push “up and to the right” to the boundary of the feasible region: the maximum occurs where both constraints hold with equality. Solve the system $\begin{cases} 2x + y = 5\\ x + 3y = 10 \end{cases}$ From the first, $y = 5 - 2x$; substituting: $x + 3(5-2x) = 10 \implies x + 15 - 6x = 10 \implies -5x = -5 \implies x = 1$, so $y = 3$.

There $x + y = 1 + 3 = 4$. Any move toward larger values would violate at least one constraint, so $4$ is the maximum. Check: $2\cdot1 + 3 = 5 \le 5$ and $1 + 3\cdot3 = 10 \le 10$. Correct. $\boxed{\max(x+y) = 4}$