This is LO

Step 1: State the linear optimization problem.

Maximize $z = 3x + 4y$

Subject to:

$\begin{cases} x + y \leq 5 \\ 2x + y \leq 8 \\ x \geq 0 \\ y \geq 0 \end{cases}$

Step 2: Graph the feasible region.

Plot lines: $x + y = 5$, $2x + y = 8$.

Intercepts:

For $x + y = 5$: $(5,0)$, $(0,5)$.

For $2x + y = 8$: $(4,0)$, $(0,8)$.

Intersection: Solve $x + y = 5$, $2x + y = 8$.

Subtract: $x = 3$, then $y = 2$.

Vertices: $(0,0)$, $(4,0)$, $(3,2)$, $(0,5)$.

Step 3: Evaluate objective at vertices.

At $(0,0)$: $z = 3(0) + 4(0) = 0$.

At $(4,0)$: $z = 3(4) + 4(0) = 12$.

At $(3,2)$: $z = 3(3) + 4(2) = 9 + 8 = 17$.

At $(0,5)$: $z = 3(0) + 4(5) = 20$.

Step 4: Maximum value.

Maximum $z = 20$ at $(0,5)$.

Final answer: Maximum value is $20$ at $x=0$, $y=5$.

$\boxed{z = 20 \quad (x=0, y=5)}$