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+// Example 35_4
+clc;funcprot(0);
+//Given data
+// C_1=100*10^4+ 600 kW+0.1 kWh
+// C_2=60*10^4+ 350 kW+0.2 kWh
+A_1=600;
+A_2=350;
+B_1=0.1;
+B_2=0.2;
+t=8760;// hours
+
+// Calculation
+// C=[100*10^4+600x+0.1S_a]+[60*10^4+350*(P-x)+0.2*(S_t-S_a)];
+//  The required condition is dC/dx=0;
+// dS_a=2500dx;....(a)
+// From Fig.Prob.35.4(b)
+// dS=H.dx;....(b)
+H=2500;// hrs
+// From similar triangles oab and dcb
+ob=50000;// kW
+L_b=ob;// kW
+db=(H*ob)/t;// Installed capacity for station B in kW
+S_b=(1/2)*db*H;// Units generated by station B
+oa=t;// hours
+S_t=(1/2)*ob*oa;// Total units generated in kWh
+S_a=S_t-S_b;// Units generated by station A
+L_a=ob-db;// kW
+C_a=100*10^4+(A_1*L_a)+(0.1*S_a);// rupees
+C_b=60*10^4+(A_2*ob)+(0.1*S_b);// rupees
+C=C_a+C_b;// rupees
+Gc=(C/S_t)*100;// Overall cost of generation in paise/kWh
+printf('\nOverall cost of generation=%0.1f paise/kWh',Gc);
+// The answer vary due to round off error