1. **State the problem:** We want to maximize the objective function $$P = 8x + 8y$$ subject to the constraints: $$2x + y \leq 10$$ $$x + 2y \leq 8$$ $$x \geq 0, y \geq 0$$ 2. **Identify the feasible region:** The feasible region is the set of all points $(x,y)$ that satisfy all constraints simultaneously. 3. **Find the corner points of the feasible region:** - From $2x + y \leq 10$, when $x=0$, $y \leq 10$; when $y=0$, $x \leq 5$. - From $x + 2y \leq 8$, when $x=0$, $y \leq 4$; when $y=0$, $x \leq 8$. - Also, $x,y \geq 0$ restricts us to the first quadrant. 4. **Calculate intersection points of the constraints:** Solve the system: $$\begin{cases} 2x + y = 10 \\ x + 2y = 8 \end{cases}$$ Multiply second equation by 2: $$2x + 4y = 16$$ Subtract first equation: $$2x + 4y - (2x + y) = 16 - 10$$ $$3y = 6 \implies y = 2$$ Substitute $y=2$ into $2x + y = 10$: $$2x + 2 = 10 \implies 2x = 8 \implies x = 4$$ 5. **List corner points:** - $(0,0)$ - $(0,4)$ from $x+2y=8$ - $(5,0)$ from $2x+y=10$ - $(4,2)$ intersection point 6. **Evaluate $P$ at each corner point:** - At $(0,0)$: $P = 8(0) + 8(0) = 0$ - At $(0,4)$: $P = 8(0) + 8(4) = 32$ - At $(5,0)$: $P = 8(5) + 8(0) = 40$ - At $(4,2)$: $P = 8(4) + 8(2) = 32 + 16 = 48$ 7. **Conclusion:** The maximum value of $P$ is **48** at the point $(4,2)$.