1. **Problem Statement:** Maximize profit $P = x + 2y$ subject to constraints: $$x + y \leq 5$$ $$2 + 3y \leq 12$$ $$x, y \geq 0$$ 2. **Formulas and Rules:** Linear programming problems maximize or minimize a linear objective function subject to linear inequalities. The feasible region is the intersection of all constraints. The maximum or minimum occurs at vertices (corner points) of the feasible region. 3. **Rewrite constraints:** From $2 + 3y \leq 12$, subtract 2: $$3y \leq 10 \implies y \leq \frac{10}{3} \approx 3.33$$ 4. **Find intersection points of constraints:** - Intersection of $x + y = 5$ and $y = \frac{10}{3}$: $$x + \frac{10}{3} = 5 \implies x = 5 - \frac{10}{3} = \frac{15}{3} - \frac{10}{3} = \frac{5}{3} \approx 1.67$$ - Intersection of $x + y = 5$ and $x=0$: $$0 + y = 5 \implies y=5$$ but $y \leq \frac{10}{3}$ so max $y$ is $3.33$ here. - Intersection of $y = \frac{10}{3}$ and $x=0$: $$x=0, y=3.33$$ 5. **Vertices of feasible region:** - $(0,0)$ from $x,y \geq 0$ - $(0, \frac{10}{3})$ - $(\frac{5}{3}, \frac{10}{3})$ - $(5,0)$ from $x + y = 5$ and $y=0$ 6. **Evaluate profit $P = x + 2y$ at vertices:** - At $(0,0)$: $P=0 + 2\times0=0$ - At $(0, \frac{10}{3})$: $P=0 + 2 \times \frac{10}{3} = \frac{20}{3} \approx 6.67$ - At $(\frac{5}{3}, \frac{10}{3})$: $P= \frac{5}{3} + 2 \times \frac{10}{3} = \frac{5}{3} + \frac{20}{3} = \frac{25}{3} \approx 8.33$ - At $(5,0)$: $P=5 + 0 = 5$ 7. **Maximum profit:** Occurs at $(\frac{5}{3}, \frac{10}{3})$ with $P = \frac{25}{3} \approx 8.33$ --- **Summary:** - Feasible region is bounded by $x + y \leq 5$, $y \leq \frac{10}{3}$, and $x,y \geq 0$. - Maximum profit is approximately 8.33 at $x=1.67$, $y=3.33$.