1. **State the problem:** The farm owner has 140 acres of land to plant corn and soybeans. Each acre of corn yields a profit of 370 and requires 2 hours of labor. Each acre of soybeans yields a profit of 290 and requires 1 hour of labor. The total labor available is 320 hours. We want to find how many acres of corn ($x$) and soybeans ($y$) to plant to maximize profit. 2. **Define variables:** Let $x$ = acres of corn Let $y$ = acres of soybeans 3. **Write constraints:** - Land constraint: $$x + y \leq 140$$ - Labor constraint: $$2x + y \leq 320$$ - Non-negativity: $$x \geq 0, y \geq 0$$ 4. **Profit function to maximize:** $$P = 370x + 290y$$ 5. **Find corner points of the feasible region:** - Intersection of land and labor constraints: Solve system: $$x + y = 140$$ $$2x + y = 320$$ Subtract first from second: $$2x + y - (x + y) = 320 - 140 \Rightarrow x = 180$$ Then $$y = 140 - 180 = -40$$ (not feasible since $y \geq 0$) - Check intercepts: For land constraint: $$x=0 \Rightarrow y=140$$ $$y=0 \Rightarrow x=140$$ For labor constraint: $$x=0 \Rightarrow y=320$$ (not feasible since land max is 140) $$y=0 \Rightarrow 2x=320 \Rightarrow x=160$$ (not feasible since land max is 140) 6. **Check feasible corner points:** - $(0,0)$ - $(0,140)$ (land max) - $(140,0)$ (land max) - Intersection of labor constraint with $y=0$ is $x=160$ but land max is 140, so max $x=140$ - Intersection of labor constraint with $x=0$ is $y=320$ but land max is 140, so max $y=140$ 7. **Check labor constraint at $(140,0)$:** $$2(140) + 0 = 280 \leq 320$$ feasible Check labor constraint at $(0,140)$: $$2(0) + 140 = 140 \leq 320$$ feasible 8. **Check intersection of labor constraint and land constraint at $y=0$ or $x=0$ is outside land limit, so check intersection of labor constraint and land constraint at $y=140 - x$:** Substitute $y=140 - x$ into labor constraint: $$2x + (140 - x) \leq 320 \Rightarrow x + 140 \leq 320 \Rightarrow x \leq 180$$ Since land max is 140, $x=140$ is feasible. 9. **Evaluate profit at corner points:** - At $(0,140)$: $$P = 370(0) + 290(140) = 40600$$ - At $(140,0)$: $$P = 370(140) + 290(0) = 51800$$ - At $(100,40)$ (check if feasible): Check labor: $$2(100) + 40 = 240 \leq 320$$ Check land: $$100 + 40 = 140$$ Profit: $$370(100) + 290(40) = 37000 + 11600 = 48600$$ 10. **Maximum profit is at $(140,0)$ with profit 51800.** **Final answer:** The farm owner should plant 140 acres of corn and 0 acres of soybeans to yield a maximum profit of 51800.