# GATE 2015 EE – SET 1 – Question 41

Solution :

$$\frac{dC_{1}}{dP_{1}}$$ = $$\frac{dC_{2}}{dP_{2}}$$

P_{1}+P{2}= 200MW ———–(1)

$$\frac{dC_{1}}{dP_{1}}$$ = $$\frac{dC_{2}}{dP_{2}}$$

(0.01)(2)$$P_{1}$$+30 = (0.05)(2)$$P_{2}$$+10

0.02$$P_{1}$$ + 30 = 0.1$$P_{2}$$+ 10

0.1 $$P_{2}$$- 0.02 $$P_{1}$$ = 20 ——–(2)

Solving (1) and (2)

$$P_{1}$$=0 ,$$P_{2}$$ =200

$$P_{1}$$ = 100MW , $$P_{2}$$ = 100 MW

$$\frac{dC_{2}}{dP_{2}}$$ = 0.1 $$P_{2}$$ + 10 = 0.1(100)+10 = 10 + 10 = 20