Contents
Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
%
E7 Summer 2010 Assignment 6 Solutions
%
Lecturer George Anwar
%
created 7/2010, Kenny Lee (reader)
Problem 1
Q = [0:0.01:8]*1e6;
% Sets independent variable array
revenue = ((6e6-Q)/1.1e6).*Q;
% 'P' is revenue PER UNIT 'Q'
cost = 2.045e6+(0.62+0.24+0.16)*Q;
% Cost including 'fixed','material','energy','labor'
figure
% Opens a new figure
plot(Q*1e-6,revenue*1e-6,
...
% Plot quantity (in units of millions) vs. revenue/cost (in units of mill
Q*1e-6,cost*1e-6); grid
on
% Displays gridlines
title({
'E7 Assignment 6 Problem 1'
;
...
% '...' syntax allows you to start a new line. MATLAB ignores this, just
'Revenue and Cost of Chemical Product'
});
% Titles the plot with two lines
xlabel(
'Q (millions of gallons)'
);
% Labels x-axis
ylabel(
'Revenue / Cost (millions of dollars)'
);
% Labels y-axishold on;
legend(
'Revenue'
,
'Cost'
);
% Legend
gtext(
'Breakeven Point 1'
),
...
gtext(
'Breakeven Point 2'
);
% Labels with user selection. TEXT() can be used as well.
%
Breakeven points are approximately at Q=0.5 and Q=4.4 million gallons.
%
Any value of Q between these values will be profitable. Producing less
%
than 0.5 million gallons will be unprofitable because of the fixed costs.
%
Producing more than 4.4 million gallons will be unprofitable because the
%
selling price will drop too much.
%
Profit is calculated as the difference between revenue and cost.
figure
plot(Q,revenue-cost); grid
on
title({
'E7 Assignment 6 Problem 1'
;
...
'Profit of Selling Chemical Product'
});
xlabel(
'Q (millions of gallons)'
);
ylabel(
'Profit (millions of dollars)'
);
%
Maximum profit occurs approximately with 2.4 million gallons of product.

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Problem 2
Guess and check to home in on the roots. Example of axes is shown below:
figure
fplot(
'x*tan(x)-9'
,[0 50]); grid
on
% Plots the equation with x-axis range defined by [] syntax
title({
'E7 Assignment 6 Problem 2'
;
...
'Zoomed Out View of the First 16 Roots of y=x*tan(x)-9'
});
xlabel(
'Independent Variable x'
);
ylabel(
'Dependent Variable y'
);
figure
fplot(
'x*tan(x)-9'
,[1.15 1.5]); grid
on
title({
'E7 Assignment 6 Problem 2'
;
...
'Zoomed In View of the 1st Root of y=x*tan(x)-9'
});
xlabel(
'Independent Variable x'
);
ylabel(
'Dependent Variable y'
);
figure
fplot(
'x*tan(x)-9'
,[4 4.5]); grid
on
title({
'E7 Assignment 6 Problem 2'
;
...
'Zoomed In View of the 2nd Root of y=x*tan(x)-9'
});
xlabel(
'Independent Variable x'
);
ylabel(
'Dependent Variable y'
);
figure
fplot(
'x*tan(x)-9'
,[6.9 7.4]); grid
on
title({
'E7 Assignment 6 Problem 2'
;
...
'Zoomed In View of the 3rd Root of y=x*tan(x)-9'
});
xlabel(
'Independent Variable x'
);
ylabel(
'Dependent Variable y'
);
%
The first plot shows 16 roots for 0<=x<=50, so there probably are an
%
infinite number of positive roots. The first three roots were found with
%
the above session: x=1.41,4.27,7.18.