1. **Stating the problem:** We have two products: boxes of strawberries ($x$) and boxes of raspberries ($y$). We want to maximize profit under resource and production constraints. 2. **Define variables:** $x$ = number of strawberry boxes $y$ = number of raspberry boxes 3. **Write inequalities for constraints:** - Water: each strawberry box needs 2 units, raspberry box 1 unit, total water ≤ 24 $$2x + y \leq 24$$ - Labor: strawberry box 1 hour, raspberry box 2 hours, total labor ≤ 24 $$x + 2y \leq 24$$ - Minimum production: $$x \geq 2$$ $$y \geq 2$$ - Storage limit for strawberries: $$x \leq 10$$ 4. **Region admissible inequalities:** $$\begin{cases} 2x + y \leq 24 \\ x + 2y \leq 24 \\ x \geq 2 \\ y \geq 2 \\ x \leq 10 \end{cases}$$ 5. **Function objective (profit):** $$P = 50x + 40y$$ 6. **Find vertices of the feasible region:** Vertices are intersections of the boundary lines within constraints. - Intersection of $2x + y = 24$ and $x + 2y = 24$: Solve system: $$2x + y = 24$$ $$x + 2y = 24$$ Multiply second by 2: $$2x + 4y = 48$$ Subtract first: $$2x + 4y - (2x + y) = 48 - 24 \Rightarrow 3y = 24 \Rightarrow y = 8$$ Substitute back: $$2x + 8 = 24 \Rightarrow 2x = 16 \Rightarrow x = 8$$ Vertex: $(8,8)$ - Intersection of $2x + y = 24$ and $y = 2$: Substitute $y=2$: $$2x + 2 = 24 \Rightarrow 2x = 22 \Rightarrow x = 11$$ But $x \leq 10$, so limit $x=10$, $y=2$ is vertex. - Intersection of $x = 10$ and $x + 2y = 24$: Substitute $x=10$: $$10 + 2y = 24 \Rightarrow 2y = 14 \Rightarrow y = 7$$ Vertex: $(10,7)$ - Intersection of $x = 2$ and $x + 2y = 24$: Substitute $x=2$: $$2 + 2y = 24 \Rightarrow 2y = 22 \Rightarrow y = 11$$ But $y \geq 2$ and no upper limit, so vertex $(2,11)$ - Intersection of $x=2$ and $y=2$: Vertex: $(2,2)$ - Intersection of $x=10$ and $y=2$: Vertex: $(10,2)$ 7. **Check which vertices satisfy all constraints:** - $(8,8)$: check $x \leq 10$, $y \geq 2$, $x \geq 2$, $y \geq 2$ all true. - $(10,2)$: all constraints true. - $(10,7)$: all constraints true. - $(2,11)$: all constraints true. - $(2,2)$: all constraints true. - $(11,2)$ invalid since $x \leq 10$. 8. **Calculate profit at each vertex:** - $P(8,8) = 50\times8 + 40\times8 = 400 + 320 = 720$ - $P(10,2) = 50\times10 + 40\times2 = 500 + 80 = 580$ - $P(10,7) = 50\times10 + 40\times7 = 500 + 280 = 780$ - $P(2,11) = 50\times2 + 40\times11 = 100 + 440 = 540$ - $P(2,2) = 50\times2 + 40\times2 = 100 + 80 = 180$ 9. **Maximum profit is $780$ at $(10,7)$**. **Final answer:** Produce 10 boxes of strawberries and 7 boxes of raspberries to maximize profit.