1. **Stating the problem:** We want to maximize the power output function $$Z = 400X + 300Y$$ where $$X$$ and $$Y$$ represent the number of Monocrystalline and Polycrystalline solar panels respectively. 2. **Constraints:** The system is subject to the following constraints: - $$2X + Y \leq 20$$ - $$2X + 3Y \leq 50$$ - $$X, Y \geq 0$$ 3. **Understanding the constraints:** These inequalities define a feasible region where the solution must lie. The first constraint limits the combination of panels based on one resource, and the second constraint limits based on another. 4. **Finding intersection points:** To find the vertices of the feasible region, solve the system: $$\begin{cases} 2X + Y = 20 \\ 2X + 3Y = 50 \end{cases}$$ Subtracting the first from the second: $$2X + 3Y - (2X + Y) = 50 - 20 \Rightarrow 2Y = 30 \Rightarrow Y = 15$$ Substitute $$Y=15$$ into $$2X + Y = 20$$: $$2X + 15 = 20 \Rightarrow 2X = 5 \Rightarrow X = 2.5$$ 5. **Evaluating the objective function at vertices:** Vertices include: - $$A(0,0)$$: $$Z = 400(0) + 300(0) = 0$$ - $$B(0, \frac{50}{3} \approx 16.67)$$: Check if feasible, but since $$Y=16.67$$ violates $$2X + Y \leq 20$$ at $$X=0$$, it's outside feasible region. - $$C(2,15)$$ (integer solution near intersection): $$Z = 400(2) + 300(15) = 800 + 4500 = 5300$$ - $$D(10,0)$$ (from $$2X + Y = 20$$ when $$Y=0$$): $$Z = 400(10) + 300(0) = 4000$$ 6. **Optimal solution:** The integer solution $$X=2$$, $$Y=15$$ yields the maximum power output $$Z=5300$$ watts within the constraints. **Final answer:** $$\boxed{X=2, Y=15, Z=5300}$$ watts is the optimal integer solution maximizing power output under given constraints.