Contents

Aaron Klapheck

% Ch 5. Assignment #11       due 11/21
clear, clc
Date = date

Date =

12-Dec-2007

Problem 2

% GIVEN

% Costs:
% Fixed (in $ per year)
%   2045000
% variable (in $ per gallon of product)
%   material -  0.62
%   energy   -  0.24
%   labor    -  0.16
% Total_cost ($) = fixed + (material+energy+labor)*Q

% Q = 6*10^6 - 1.1*10^6*P , where P is selling price ($ per gallon),
% and Q is the sales quantity (gallons).
% Solve for P: P = (6*10^6 - Q)/1.1*10^6
% P*Q = ($/gallon)*(gallons) = Revenue ($) = ((6*10^6 -Q)/(1.1*10^6))*Q
% Note: profit = revenue - cost.

% PLOT

% Plot Total_cost vs. Q and Revenue vs. Q.

% Given
material = 0.62; energy = 0.24; labor = 0.16; fixed = 2045000;
% Set up range of Q.
Q = [0:100:10^7];
Total_cost = fixed + (material+energy+labor)*Q;
Revenue = ((6*10^6 -Q)/(1.1*10^6)).*Q;

% General plot
% The purpose of this plot is to get an idea of what the graph looks like.

plot(Q, Total_cost, Q, Revenue), grid, xlabel('Q'), ...
    title('Total Cost and Revenue vs. Sales Quantity'), ...
    legend('Cost($) vs. Q(gal)', 'Revenue($) vs. Q(gal)'), ...
    ylabel('Cost and Revenue')

Problem 2 - fist break even point (the one on the left)

plot(Q, Total_cost, Q, Revenue), grid, xlabel('Q'),  ...
    title('Total Cost and Revenue vs. Sales Quantity'), ...
    legend('Cost($) vs. Q(gal)', 'Revenue($) vs. Q(gal)'), ...
    ylabel('Cost and Revenue'),
    axis([5*10^5 5.3*10^5 2.5*10^6 2.6*10^6]), ...
    [Q_1, Revenue_1] = ginput(1)

Q_1 =

  5.1573e+005


Revenue_1 =

  2.5709e+006

Problem 2 - second break even point (the one on the right)

plot(Q, Total_cost, Q, Revenue), grid, xlabel('Q'),  ...
    title('Total Cost and Revenue vs. Sales Quantity'), ...
    legend('Cost($) vs. Q(gal)', 'Revenue($) vs. Q(gal)'), ...
    ylabel('Cost and Revenue'),
    axis([4.3*10^6 4.4*10^6 6.4*10^6 6.6*10^6]), ...
    [Q_2, Revenue_2] = ginput(1)

Q_2 =

  4.3623e+006


Revenue_2 =

  6.4944e+006

Problem 2 - maximum profit and state range

plot(Q, Total_cost, Q, Revenue), grid, xlabel('Q'),  ...
    title('Total Cost and Revenue vs. Sales Quantity'), ...
    legend('Cost($) vs. Q(gal)', 'Revenue($) vs. Q(gal)'), ...
    ylabel('Cost and Revenue'),
    axis([2.9*10^6 3.1*10^6 7*10^6 10*10^6]), ...
    [Q_max, Revenue_max] = ginput(1)

% Range of Q is from Q_1 to Q_2
Range_of_Q_for_profitablility = [Q_1, Q_2]

Q_max =

  3.0002e+006


Revenue_max =

  8.1798e+006


Range_of_Q_for_profitablility =

  1.0e+006 *

    0.5157    4.3623

Problem 3

% Graphicaly find the roots of the given polynomial.


x = [-10:0.01:10];
p = [4, 3, -95, 5, -10, 80];

% General plot
plot(x, polyval(p,x)), title('Finding Roots of a Polynomial'), grid, ...
    xlabel('x'), ylabel('y = 4x^5+3x^4-95x^3+5x^2-10x+8')

Problem 3 - Use plot to find first root

plot(x, polyval(p,x)), title('Finding Roots of a Polynomial'), grid, ...
    xlabel('x'), ylabel('y = 4x^5+3x^4-95x^3+5x^2-10x+8'), ...
    axis([-5.4 -5.2 -0.1 0.1]), [x1_root, y1] = ginput(1)

x1_root =

   -5.3094


y1 =

 -2.9240e-004

Problem 3 - Use plot to find second root

plot(x, polyval(p,x)), title('Finding Roots of a Polynomial'), grid, ...
    xlabel('x'), ylabel('y = 4x^5+3x^4-95x^3+5x^2-10x+8'), ...
    axis([0.8 1 -0.1 0.1]), [x2_root, y2] = ginput(1)

x2_root =

    0.9459


y2 =

 -8.7719e-004

Problem 3 - Use plot to find third root and double check

plot(x, polyval(p,x)), title('Finding Roots of a Polynomial'), grid, ...
    xlabel('x'), ylabel('y = 4x^5+3x^4-95x^3+5x^2-10x+8'), ...
    axis([4.4 4.5 -0.1 0.1]), [x3_root, y3] = ginput(1)


% Using graphs: found roots to be at the x values: -5.3094, 0.946, and 4.474.

% Double check
Actual_root_of_polynomial = roots(p)
Graphical_root_of_polynomial = [x1_root, x2_root, x3_root]

x3_root =

    4.4741


y3 =

 -2.9240e-004


Actual_root_of_polynomial =

  -5.3094          
   4.4740          
   0.9461          
  -0.4303 + 0.8395i
  -0.4303 - 0.8395i


Graphical_root_of_polynomial =

   -5.3094    0.9459    4.4741

Problem 6

% Given
% V(t) = 10^9 + 10^8*(1 - exp(-t/100)) - 10^7*t, Where
% V is the water volume in liters
% t is the time in days

% 50% of 10^9 = 5*10^8

% Plot V(t) vs. t.

t = [0:1:100];
V = 10^9 + 10^8*(1 - exp(-t/100)) - 10^7*t;
plot(t, V), axis([0 100 4.9*10^8 5.1*10^8]), ...
    Title('Water Volume in Reservoir vs. Time'), ...
    xlabel('Time (Days)'), ylabel('Water Volume (Liters)'), grid, ...
    [days, liters_of_water] = ginput(1)

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days =

   54.4931


liters_of_water =

  4.9991e+008

Problem 9

% Given:
A = [0, -8, 6; 5, -4, 3; 10, -1, 1; 15, 1, 0; 20, 2, -1]

% plot column 1 vs. column 2 and 3 of matrix A.
plot(A(:, 1),A(:, 2:3)), Title('Forces Applied for a Given Time'), ...
    xlabel('Time (s)'), ylabel('Force (N)')

A =

     0    -8     6
     5    -4     3
    10    -1     1
    15     1     0
    20     2    -1

Problem 11

x = [0:0.01:2*pi];
y = tan(2*x);
subplot(1,2,1)
plot(x, y), title('tan(2x) vs. x'), xlabel('x'), ylabel('tan(2x)')


x = [0:0.01:2*pi];
y = (2*tan(x))./(1-tan(x).^2);
subplot(1,2,2)
plot(x, y), title('2tan(x)/(1-tan(x)^2) vs. x'), xlabel('x'), ...
    ylabel('2tan(x)/(1-tan(x)^2)')

% They look identical.

Problem 14

% Steady state value of y is 1. 98% of 1 is 0.98, where y is defined as:
% y = 1 - exp(-bt) equation #1

t = [0:0.01:30];
c = [1:1:20];

y_98 = 0.98

b_1 = c(1);
y_1 = 1 - exp(-b_1.*t);

b_10 = c(10);
y_10 = 1 - exp(-b_10.*t);

b_2 = c(2);
y_2 = 1 - exp(-b_2.*t);

subplot(1,1,1)
plot(t, y_1, t, y_2, t, y_10, t, y_98, '--'), axis([0 6 0 2]), ...
    title('(1 - exp(-bt)) vs. t at Various Values of b'), ...
    xlabel('Time'), ylabel('1 - exp(-bt)'), grid, ...
    legend('1 - exp(-t)', '1 - exp(-10t)', '1 - exp(-20t)', 'y = 0.98')

% Used graph to get the following information for equation #1.
% When b = 1,  98% of y occurs at x = 4
% When b = 2,  98% of y occurs at x = 2
% When b = 10,  98% of y occurs at x = 0.4

y_98 =

    0.9800

Problem 15

t = [0:0.01:10];
x = 10.*exp((-0.5).*t).*sin(3.*t + 2);
y = 7.*exp((-0.4).*t).*cos(5.*t - 3);

plot(t, x, t, y), xlabel('Time'), ylabel('Oscillations and Vibrations'), ...
    title('Electical Oscillations and Machine Vibrations vs. Time'), ...
    legend('Electrical Circuit Oscillations', 'Machine Vibrations')


Published with MATLAB® 7.0.1