1. **Nyatakan masalah:** Diberikan fungsi biaya bahan bakar tiga stasiun pembangkit thermal: $$C_1 = 600 + 5.3P_1 + 0.005P_1^2$$ $$C_2 = 400 + 5.5P_2 + 0.007P_2^2$$ $$C_3 = 300 + 5.8P_3 + 0.008P_3^2$$ Dengan batasan daya: $$200 \leq P_1 \leq 450$$ $$150 \leq P_2 \leq 350$$ $$100 \leq P_3 \leq 225$$ Dan beban total: $$P_1 + P_2 + P_3 = 1000$$ 2. **Tujuan:** Menentukan nilai $P_1$, $P_2$, dan $P_3$ yang meminimalkan total biaya: $$C = C_1 + C_2 + C_3$$ 3. **Metode:** Gunakan metode Lagrange multiplier dengan fungsi: $$L = C_1 + C_2 + C_3 - \lambda (P_1 + P_2 + P_3 - 1000)$$ 4. **Turunkan persamaan untuk tiap variabel:** $$\frac{\partial L}{\partial P_1} = 5.3 + 0.01P_1 - \lambda = 0$$ $$\frac{\partial L}{\partial P_2} = 5.5 + 0.014P_2 - \lambda = 0$$ $$\frac{\partial L}{\partial P_3} = 5.8 + 0.016P_3 - \lambda = 0$$ 5. **Ekspresikan $P_1$, $P_2$, dan $P_3$ dalam fungsi $\lambda$:** $$P_1 = 100(\lambda - 5.3)$$ $$P_2 = \frac{100}{1.4}(\lambda - 5.5) \approx 71.43(\lambda - 5.5)$$ $$P_3 = 62.5(\lambda - 5.8)$$ 6. **Gunakan batasan total daya:** $$P_1 + P_2 + P_3 = 1000$$ Substitusi: $$100(\lambda - 5.3) + 71.43(\lambda - 5.5) + 62.5(\lambda - 5.8) = 1000$$ 7. **Selesaikan persamaan:** $$100\lambda - 530 + 71.43\lambda - 392.865 + 62.5\lambda - 362.5 = 1000$$ $$234.93\lambda - 1285.365 = 1000$$ $$234.93\lambda = 2285.365$$ $$\lambda \approx 9.73$$ 8. **Hitung $P_1$, $P_2$, dan $P_3$:** $$P_1 = 100(9.73 - 5.3) = 443$$ $$P_2 = 71.43(9.73 - 5.5) = 305.7$$ $$P_3 = 62.5(9.73 - 5.8) = 247.5$$ 9. **Periksa batasan:** $P_3 = 247.5$ melebihi batas maksimum 225, jadi set $P_3 = 225$ dan hitung ulang $P_1$ dan $P_2$ dengan: $$P_1 + P_2 = 1000 - 225 = 775$$ 10. **Hitung ulang dengan $P_3=225$:** Dari turunan $\frac{\partial L}{\partial P_3} = 0$: $$\lambda = 5.8 + 0.016 \times 225 = 5.8 + 3.6 = 9.4$$ Substitusi $\lambda=9.4$ ke $P_1$ dan $P_2$: $$P_1 = 100(9.4 - 5.3) = 410$$ $$P_2 = 71.43(9.4 - 5.5) = 279.3$$ 11. **Periksa batasan:** $P_1=410$ (antara 200 dan 450), $P_2=279.3$ (antara 150 dan 350), $P_3=225$ (batas atas), valid. 12. **Hitung total biaya:** $$C_1 = 600 + 5.3 \times 410 + 0.005 \times 410^2 = 600 + 2173 + 840.5 = 3613.5$$ $$C_2 = 400 + 5.5 \times 279.3 + 0.007 \times 279.3^2 = 400 + 1536.15 + 545.3 = 2481.45$$ $$C_3 = 300 + 5.8 \times 225 + 0.008 \times 225^2 = 300 + 1305 + 405 = 2010$$ $$C_{total} = 3613.5 + 2481.45 + 2010 = 8105$$ **Jawaban akhir:** Daya optimal: $$P_1 = 410 \text{ MW}, P_2 = 279.3 \text{ MW}, P_3 = 225 \text{ MW}$$ Biaya total bahan bakar: $$C_{total} = 8105 \text{ per jam}$$