Exercise 5.8: Solve the following diet problem:
| X1 X2 | Min. Req.
--------+----------+------------
E1 | 10 4 | 40
E2 | 7 7 | 49
E3 | 5 10 | 50
--------+----------+------------
Price | 5 7 |
We could have used the tableau method like in the last
exercise, but we will show the alternative graphical method.
The plot is made with MATLAB.
>> e1 = [-10/4 10]
e1 =
-2.5000 10.0000
>> e2 = [-1 7]
e2 =
-1 7
>> e3 = [-1/2 5]
e3 =
-0.5000 5.0000
>> r = -0.1:0.1:4.2; % plot range
>> y1 = polyval(e1,r);
>> y2 = polyval(e2,r);
>> y3 = polyval(e3,r);
>> figure
>> hold on
>> plot(r,y1,'b');
>> plot(r,y2,'r');
>> plot(r,y3,'g');
>> p6 = [-5/7 6];
>> yp6 = polyval(p6,r);
>> plot(r,yp6,'black');
>> p5 = [-5/7 5];
>> yp5 = polyval(p5,r);
>> plot(r,yp5,'black');
% Notice that the intersection of the first and third constraint
% will not satisfy the second constraint. From the slope of the
% objective function, we can see that the optimal value occurs
% at the intersection of the second and third constraint:
>> a = [7 7; 5 10]
a =
7 7
5 10
>> b = [49; 50]
b =
49
50
>> x = a\b
x =
4
3
>> % computing the value of the objective function:
>> 5*x(1) + 7*x(2)
ans =
41
Here is the plot:
