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Mathematical Computing
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 
Numerical Computing in Matlab
% The code you used to get the answer: n = 10; digitsWhole = digits(n+(numel(num2str(floor(pi))))); MyPi = vpa(pi); digits_after =char(rem(MyPi,1)); % 10 digits as required ref = strfind(digits_after,'.'); tenth_digit = str2num(digits_after(ref+n)); % The 10th digit of pi after the decimal is: fprintf('The 10th digit of pi after the decimal is: %g \n' ,tenth_digit) % The value of y didn't appear in the command window because: % *It was suppressed with a semicolon (;)* % Your code: horizontal_vec = [8 2 6 10] % Your code: vertical_vec = [5; 9; -9; 0] %OR: vertical_vec = [5 9 -9 0]' % Your code: k=1; for i=98:-1:42 if mod(i,2)~=0 && i~=1 odd(k)=i; k=k+1; end end odd % expected horizontal vector num2str(odd) % Your code: k=1; for i=5:1:101 if mod(i,2)==0 && i~=1 even(k)=i; k=k+1; end end even % expected horizontal vector num2str(even) % The code you used to get the answer: X = 10:1.5:175; Number_of_element_in_X = numel(X); % The answer is: fprintf('The answer is: %g \n', Number_of_element_in_X) % Your code: n=36; a= 23; b = 27; list = linspace(a,b,n)'; num2str(list) % The code you used to create a 5 x 4 matrix of random numbers between -2 % and 5: a = -2; b = 5; m= 5; n=4; Mat_X = a + (b-a).*rand(m,n) % The code you used to determine the number in the 2nd row and 4th column: Mat_X(2,4) % The code you used to create a 3 x 7 matrix of random integers between -8 % and -2: a = -8; b = 2; m= 3; n=7; Mat_Y = randi([a,b], m, n) % The code you used to slice out the 2nd and 3rd rows: j = 1; % removes 2nd and 3rd rows Mat_Y_new = Mat_Y(j,:) % The code you used to create a 3 x 4 matrix of 0's: Mat_A = zeros(3,4); % The code you used to create a 3 x 4 matrix of 6's: Mat_B = 6* ones(3,4); % The code you used to combine them horizontally: Mat_C = horzcat(Mat_A, Mat_B) % The code you used to combine them vertically: Mat_D = vertcat(Mat_A, Mat_B) % Your code: m = 3; n = 700; for i = 1:m for j = 1:n Mat_E(i,j) = j; end end Mat_E % Alternatively: % Mat_E = [1:700].*ones(3,1) % Your code: m = 850; n = 850; for i = 1:m for j = 1:n Mat_F(j,j) = j; end end Mat_F; % Alternatively: % Mat_F = diag([1:850])