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here is linear programming question I am having trouble with. Can you show me how to solve this problem in the most easy to use fashion?

For the following Linear Programming problem answer the questions (a) to (e) below:

    Max Z = 3x + 2y

    Subject to:
    2x + y <= 8
    x + y >= 4
    y <= 10
    x, y >= 0

(a) Solve the problem to find the optimum solution. Clearly label and show the feasible solution space on your diagram.

dazzle
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  • This is not a homework site. You are supposed to show your work/thoughts on the problem to get a favourable response. – StubbornAtom Jun 25 '17 at 12:50
  • This is not a homework question - as you should know looking on the date - it is rather an examination question of the past years - I am looking for a right answer so I can recheck to know I am doing everything right. – dazzle Jun 25 '17 at 13:15
  • @dazzle Please select one of the five questions where you stuck. An answer for $a)-e)$ is impossible. – callculus42 Jun 25 '17 at 13:16
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    Thank you calculus. Answering "(a) Solve the problem to find the optimum solution. Clearly label and show the feasible solution space on your diagram." would already help a lot. What is the best approach to go about things here? Graphical method? – dazzle Jun 25 '17 at 13:30
  • I corrected the question and reduced it only to a) so its not having to many questions. – dazzle Jun 25 '17 at 13:35

1 Answers1

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First you solve the constraints for y. For the first constraint I post the step.

$2x+y\leq 8 $

Subtracting 2x on both sides.

$y\leq 8-2x \quad$ (red line)

The key point here is the $\leq$-sign. The feasible region is below the red line. Next constraint:

$x + y \geq 4$

$y\geq 4-y \quad$ (grey line)

The feasible region is above the grey line. The third constraint is

$y\leq 10 \quad$ (black line).

This constraint is not binding due to the first constraint. And finally we have the constraint $x,y\geq 0$. Thus we have to regard the first quadrant only. enter image description here

I hope it is obvious to you that the green area is the feasible space. The optimal solution can be found graphically. Let denote the value of the objective function as $z$.

$z=3x+2y \ \ $ Solving for $y$.

$z-3x=2y$

$y=\frac{z-3x}{2}$

To start we set $z$ equal to $0$. $y=-\frac{3x}{2}$

This equation can be drawn into the graph as well. It is the thick blue line. Now you push the line parallel upward till the line touches the feasible space the last time. The thin blue lines illustrate the process. The last point where the (shifted) objective function and the feasible space have a common point is $(x^*,y^*)=(0,8)$

callculus42
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