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+// Example 32_21
+clc;funcprot(0);
+//Given data
+// L=350+10t-t^2;
+
+// Calculation
+// Differentiating L with respect to t, we get 10-2t=0
+t=10/2;// hrs
+L_max=350+(10*t)-t^2;// The maximum load occurs at 5 th hour during the day in MW
+t_0=0;
+t_1=24;// Limits of integration
+L_av=(1/24)*integrate('(350+(10*t)-t^2)','t',t_0,t_1);
+F_l=L_av/L_max;// Load factor
+printf('\nMaximum load,L_max=%0.0f MW \nLoad factor of the plant=%0.4f',L_max,F_l);
+// Load duration curve is the representation of load with respect to time 
+t=[0 1 2 3 4 5  6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24];// Time in hours
+for(i=1:25)
+    L(i)=((350+(10*t(i))-t(i)^2));
+end
+T=[0 12 24];
+L_max=[L_max L_max L_max];
+subplot(2,1,1);
+plot(t',L,'g',T',L_max','--');
+xlabel('t');
+ylabel('L');
+xtitle('Load curve');
+// Load duration curve is the representation of load with respect to time is decending order.
+T=[24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0];// Time in hours
+for(j=1:25)
+    L(j)=((350+(10*T(j))-T(j)^2));
+end
+subplot(2,1,2);
+plot(t',L,'r');
+xtitle('Load duration curve');
+xlabel('t');
+ylabel('L');
+