1. The problem is to maximize the objective function $Z = 3x + 5y$ given the corner points of the feasible region: $(0,0)$, $(0,6)$, $(4,4)$, and $(6,0)$. 2. We will evaluate $Z$ at each corner point because in Linear Programming, the maximum or minimum value of the objective function occurs at one of the vertices of the feasible region. 3. Calculate $Z$ at each point: - At $(0,0)$: $Z = 3(0) + 5(0) = 0$ - At $(0,6)$: $Z = 3(0) + 5(6) = 30$ - At $(4,4)$: $Z = 3(4) + 5(4) = 12 + 20 = 32$ - At $(6,0)$: $Z = 3(6) + 5(0) = 18$ 4. Compare the values: $0, 30, 32, 18$. The maximum value is $32$ at the point $(4,4)$. 5. Therefore, the maximum value of $Z$ is $32$ at $(4,4)$.