[MOVED] Linear prog.: Maximize z = 14x + 22y subject to....

[Navyguy]

Maximize z = 14x + 22y subject to:

. . .2x + 2y > 20
. . .8x + 5y < 40
. . .x > 0, y > 0

This linear programming problem has:

. . .a. a unique solution
. . .b. multiple solutions
. . .c. unbounded solution
. . .d. no feasible solution

Well I know c. isn't right. I have run some programs on this problem but getting different answers. What I have for an answer is "b. multiple solutions". Is this right? Thanks so much for your help.

 

[galactus]

Did you graph your inequalities?. If you do, you can see why the answer is d.

 

[Navyguy]

yes i got now

http://www.egwald.com/operationsresearc ... erator.php

ran it on this site and I came up with no feasible solution also, thanks so much for checking my answer.

 

[soroban]

Re: Linear programming help pleassssssse!!!!!!!

[Hello, Navyguy!

Yes . . . by all means, make a sketch!

Maximize \(\displaystyle \,z\:=\:14x\,+\,22y\,\) subject to: \(\displaystyle \,\begin{array}{cccc}2x\,+\,2y\:\geq\:20 \\ 8x\,+\,5y\:\leq\:40\\ x\,\geq\,0 \\ y\,\geq\,0\end{array}\)
This linear programming problem has:
\(\displaystyle \;\;\)a. a unique solution\(\displaystyle \;\;\)b. multiple solutions\(\displaystyle \;\;\)c. unbounded solution\(\displaystyle \;\;\)d. no feasible solution

The last two inequalties limit us to Quadrant I.

The first inequality is: \(\displaystyle \,x\,+\,y\:\geq\:10\)
Graph the line: \(\displaystyle \,x\,+\,y\:=\:10\).
\(\displaystyle \;\;\)It has intercepts: \(\displaystyle (10,0)\) and \(\displaystyle (0,10).\)
Shade the region above it.

The second inequality is: \(\displaystyle \,8x\,+\,5y\:\leq\:40\)
Graph the line:\(\displaystyle \,8x\,+\,5y\:=\:40\)
\(\displaystyle \;\;\)It has intercepts: \(\displaystyle (5,0)\) and \(\displaystyle (0,8)\)
Shade the region below it.

We see that there is no region common to both shaded areas.

Answer: .choice (d)