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## Moffat

Master of Science, Electronic Systems & Management Engineering, Harvard University

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Calculating Derivatives
clc, clear all, close all %% Part 1 %% Question 1 syms q p C(q) = 610 - 0.2*q + 0.15*q^2; % Cost function D(p) = 260 - 20*p; % Demand function % Revenue is equal to the amount of units multiplied by the demand (price) % function R(p) = p*D(p) % The profit is equal to the Revenue minus the Cost P(p) = R(p) - subs(C(q), q, D(p)) % The marginal cost is the first derivative of the cost function MC(q) = diff(C, q) % The average cost is the Cost function divided by the total units AC(q) = C(q)/q % The marginal revenue is the first derivative of the Revenue function MR(p) = diff(R,p); %% Question 2 figure ps = 0:0.1:13; plot(ps, D(ps), 'linewidth', 2); xlabel('Price ($)'); ylabel('Demand (q)'); title('Demand vs. Price'); grid on %% Question 3 figure subplot(1,2,1) qs = 1:1:D(0); plot(qs, C(qs), 'linewidth', 2), grid on xlabel('Quantity (q)'); ylabel('Cost ($)'); title('Cost, Marginal Cost, Marginal Revenue and Average Cost vs. Quantity'); hold on plot(qs, MC(qs), 'linewidth', 2); plot(qs, AC(qs), 'linewidth', 2); plot(qs, MR(qs), 'linewidth', 2); legend('Cost', 'Marginal Cost', 'Average Cost', 'Marginal Revenue'); subplot(1,2,2) plot(qs, MC(qs), 'linewidth', 2), hold on; plot(qs, AC(qs), 'linewidth', 2); plot(qs, MR(qs), 'linewidth', 2); legend('Marginal Cost', 'Average Cost', 'Marginal Revenue'); xlabel('Quantity (q)'); ylabel('Cost ($)'); title('Marginal Cost, Average Cost and Marginal Revenue vs. Quantity'); grid on %% Question 4 % To calculate the point of elasticity of demand, we select 2 points of the % curve % For example, we selec the points (8, 100) and (9, 80) % Change in quantity q_change_perc = (9-8)/(9+8)/2 *100; p_change_perc = (80-100)/(80+100)/2 *100; p_elast_D = abs(q_change_perc/p_change_perc) % Because the value is < 1, the demand is inelastic % Now, we will calculate the intervals from P = 1 to 13, where demand is % elastic and inelastic pvec = 0:0.01:13; N = length(pvec); elast_vec = []; for i = 1:N-1 P1 = [pvec(i) D(pvec(i))]; P2 = [pvec(i+1), D(pvec(i+1))]; q_change_perc = (P2(1)-P1(1))/(P2(1)+P1(1))/2 *100; p_change_perc = (P2(2)-P1(2))/(P2(2)+P1(2))/2 *100; p_elast_D = abs(q_change_perc/p_change_perc); elast_vec = [elast_vec, p_elast_D]; end temp = pvec(find(elast_vec > 1)); elastic_interval = [temp(1), temp(end)] temp = pvec(find(elast_vec < 1)); inelastic_interval = [temp(1), temp(end)] temp = pvec(find(elast_vec == 1)); if length(temp)>0 unitary_interval = [temp(1), temp(end)] else unitary_interval = [] end %% Question 6 % The revenue is a negative parabola, so we find its vertex a = -20; b = 260; c = 0; pmax = -b/(2*a); qmax = subs(D,p,pmax); %% Question 7 % The Average Cost is a function that is inversely proportional to the % quantity % To minimize this function, we will use the MATLAB's function fmincon % We set the bounds for the quantity to: 0 < q < Inf type ObjectiveAverageCost [qmin, fval, exiflag] = fmincon('ObjectiveAverageCost', 10, [], [], [], [], 0, Inf); qmin_int = round(qmin) pmin = double(solve(D==qmin)) pmin_int = double(solve(D==qmin_int)) %% Question 8 % The profit function is a negative parabola, so the price that maximizes % it is its vertex a = -80; b = 1816; p_profit_max = -b/(2*a) q_profit_max = D(p_profit_max) %% Part 3 %% Question 10 % This time, the supply function is a price function, where the independent % variable is the quantity. % we can express the quantity in function of price as: S(p) = p-8; % The equilibrium point is where the demand and supply are the same p_equil = solve(D == S) q_equil = D(p_equil) figure ps = 0:0.1:20; plot(ps, D(ps), 'linewidth',2), hold on plot(ps, S(ps), 'linewidth', 2) plot(p_equil, D(p_equil), 'r*', 'linewidth', 2); xlabel('Price ($)'); ylabel('Quantity (q)'); title('Supply and Demand vs. Price'); legend('Demand', 'Supply'); grid on %% Question 11 % The consumer surplus is the integral between the equilibrium point and % the Demand function CS = int((D(p)-q_equil), p, 0, p_equil) %% Question 12 % The producer surplus is the integral between the Supply function and the % equilibrium point PS = int((q_equil - S(p)), p, 0, p_equil) function f = ObjectiveAverageCost(x) f = 3*x/20 + 610/x - 1/5; end 
Solving Equations Using Matlab
clc, clear all, close all %% Question 1 syms x y z %% Question 2 fprintf("\n\nQuestion 2:\n"); % let's solve using symbolic solver xsol = double(solve(x^4 + 2*x^3 - 8*x^2 - 9*x+18==0)) %% Question 3 fprintf("\n\nQuestion 3:\n"); xsol = double(solve(x^3 - 5*x^2 == -x-15)) %% Question 4 fprintf("\n\nQuestion 4:\n"); % The intersection points are the solutions of the equation sol = solve([4*x^2+5*(y-1.5)^2 == 40;y-3*x+1==0], [x,y]); xsol = double(sol.x) ysol = double(sol.y) %% Question 5 fprintf("\n\nQuestion 5:\n"); % Define the system of equations equations = [2.1*x+6.2*y-3.1*z==205; -3.7*x+10.8*y+1.1*z == -107; x+2*y-3*z==23]; sol = solve(equations, [x,y,z]); xsol = double(sol.x) ysol = double(sol.y) zsol = double(sol.z) %% Question 6 fprintf("\n\nQuestion 6:\n"); % We will use x = F1 and y = F2 equations = [(1 + cosd(30))*x + 10.3*cosd(25)*y == 0; 5*sind(30)*x - 5.6*sind(25)*y == 30]; sol = solve(equations, [x,y]); F1 = double(sol.x) F2 = double(sol.y) %% Question 7 fprintf("\n\nQuestion 7:\n"); y(x) = tan(4-3*x)-sqrt(3); % Now solve xsol = double(solve(y(x)+2*x==7, x)) %% Question 8 fprintf("\n\nQuestion 8:\n"); clear y syms y z1(x,y) = 4*x^2+5*(y-1.5)^2 -40; z2(x,y) = y-3*x+1; % The intersection points are where z1== z2 sol = solve(z1==z2, [x,y]); xsol = double(sol.x) ysol = double(sol.y) %% Question 9 fprintf("\n\nQuestion 9:\n"); % We will use x and y as our variables % Define first equation eq1 = x+y == 11; % the sum of both numbers is 11 eq2 = 10*y+x + 27 == 10*x+y; sol = solve([eq1;eq2], [x,y]); number = double(10*sol.x + sol.y) %% Question 10 fprintf("\n\nQuestion 10:\n"); d = 250; %from A to B va = 60; vb = 0; ab = -20; aa = 0; % Equation for position of a: eq1 = va*x + aa*x^2 /2; eq2 = d + vb*x + ab*x^2 /2; sol = solve(eq1 == eq2) t = double(sol); t = t(t > 0); % take only the positive value Xa = double(va*t + aa*t^2 /2) Xb = double(vb*t + ab*t^2 /2) fprintf("The trains met when they are at at distance of %.3f of station A\n", Xa)