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+// Example 34_10
+clc;funcprot(0);
+//Given data
+// dF_a/dP_a=0.065*P_a+25;
+// dF_b/dP_b=0.08*P_b+20;
+L=160;// Total load in MW
+
+// Calculation
+//(a)
+function[X]=power(y)
+    X(1)=(y(1)+y(2))-L;
+    X(2)=((0.065*y(1))+25)-((0.08*y(2))+20);
+endfunction
+y=[10 100];
+z=fsolve(y,power);
+P_a=z(1);// MW
+P_b=z(2);// MW
+//(b)
+L=160/2;//If the load is equally shared by both the units
+p_a1=P_a;
+p_a2=L;// Limits of integration
+Ic_A=integrate('((0.065*p_a)+25)','p_a',p_a1,p_a2);// Increase in cost for unit A in Rs/hr.
+p_b1=P_b;
+p_b2=L;// Limits of integration
+Ic_B=integrate('((0.08*p_b)+20)','p_b',p_b1,p_b2);// Increase in cost for unit B in Rs/hr.
+dC=Ic_A+Ic_B;
+printf('\n(a)P_a=%0.1f MW \n   P_b=%0.1f MW \n(b)The loss in fuel cost per hour=Rs.%0.0f/hr',P_a,P_b,dC);
+// The answer provided in the textbook is wrong