# Sketching feasibility area's , to satisfy inequalities

• Jul 31st 2010, 03:56 AM
gilly
Sketching feasibility area's , to satisfy inequalities
For example x<4 and y<5 i understand, but i dont understand how to draw the follwing.

y-4x < -6
2y - x >-1

all four are needed to in the feasablility area.

thanks for any help.

tom
• Jul 31st 2010, 03:58 AM
Ackbeet
Those are linear inequalities. Here's what I would do: pretend for a second that the inequality is an equality. Plot the resulting line. Then plug in a point on one side of the line to see if it satisfies the inequality. If it does, then that side of the line is admissible, and the other side is not. Make sense?
• Jul 31st 2010, 04:42 AM
gilly
OK so what your saying is that. draw the line y-4x

so if y=1 x= -4. correct?

then plug in a random point one side of the line. if it meets the inequality (-6) then that side is admissable.

hope i have interpreted correctly
• Jul 31st 2010, 04:51 AM
Prove It
No, he's saying draw the line \$\displaystyle y - 4x =- 6\$.

An alternative method...

\$\displaystyle y - 4x < -6\$

\$\displaystyle y < 4x - 6\$.

So here, draw the line \$\displaystyle y = 4x - 6\$, making sure to use a broken line, since the inequality does not have the "or equal to".

Since you are saying that \$\displaystyle y\$ is less than this line, the required region is the half-plane underneath the line. So shade the half plane above the line (what you get rid of).
• Jul 31st 2010, 05:10 AM
Ackbeet
Not quite. You need to draw the line y - 4x = -6, and the line 2y - x = -1. Solve those equations for y, if that helps.