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+// Example 34_7
+clc;funcprot(0);
+//Given data
+L_p=10;// MW
+// I=5*10^6(18+12L+0.5L^2)
+L_1=5;// MW
+L_2=7;// MW
+
+// Calculation
+//(a)
+// n=(L/I)=(1/(5*10^6((18/L)+12+0.5L)));
+// The efficiency will be maximum when ((18/L)+12+0.5L)),differentiating we get
+L_m=sqrt((-18)/(-0.5));// MW
+L=L_m*10^3*3600;// kJ/hr
+I_6=5*10^6*(18+(12*L_m)+(0.5*L_m^2));// kJ/hr
+n_max=(L/I_6)*100;// Maximum efficiency in %
+printf('\n(a)The load at which the efficiency of the plant will be maximum=%0.0f MW \n   The maximum efficiency=%0.0f percentage',L_m,n_max);
+//(b)
+I_5=5*10^6*(18+(12*L_1)+(0.5*L_1^2));// kJ/hr
+I_7=5*10^6*(18+(12*L_2)+(0.5*L_2^2));// kJ/hr
+dI=I_7-I_5;// Increase in output to the plant per hour in  kJ/hr
+L=(L_1+L_2)/2;// MW
+IR=5*10^6*(12+L);// kJ/hr
+Ti=IR*(L_2-L_1);// Total increase in input in kJ/hr
+printf('\n(b)Total increase in input=%0.2e kJ/hr',Ti);
+