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+clear ;
+clc;
+// Example 3.15
+printf('Example 3.15\n\n');
+printf('Page No. 77\n\n');
+
+// given
+i_t = [20 40 60 80 100];// Insulation thickness in mm
+f_c = [2.2 3.5 4.8 6.1 7.4];// Fixed costs in (10^3 Pound / year)
+h_c = [10.2 6.5 5.2 4.6 4.2];// Heat costs in (10^3 Pound / year)
+t_c = [12.4 10 10 10.7 11.6];// Total costs in (10^3 Pound / year)
+
+//(a) Graphical solution
+//Refer figure 3.8
+C_T = 9750;// Minimum total cost in Pound
+t = 47;// Corresponding thickness of insulation in mm
+printf('The most economic thickness of insulation is %.0f mm \n',t)
+
+//(b) Numerical solution
+// The cost due to heat losses,C1, and the fixed costs,C2, vary according to the equations;-
+// C1 = (a/x) + b     and    C2 = (c*x) + d
+// Substituting the values of C1 and C2 together with the corresponding insulation thickness values , the following equations are obtained :-
+// C1 = (150*10^3/x) + 2.7*10^3    and    C2 = (65*x) + 0.9*10^3
+//And to obtain the total costs
+//CT = C1 + C2 = (150*10^3/x) + (65*x) + 3.6*10^3
+// Differentiate to optimise, and put dCT/dx equal to zero
+//dCT/dx =-((150*10^3)/x^2) + 65 = 0
+
+//Let y = dCT/dx
+function y=fsol1(x)
+  y = -((150*10^3)/x^2) + 65;
+endfunction
+[xres]=fsolve(50,fsol1);
+x = xres;
+printf('The optimum thickness of insulation is %.0f mm \n',x)