What are the vertices of these contstraints if they were graphed? 30x+40y≤2100, 600x+900y≤45000, x≥0, y≥0
And how would I graph them?

Question

What are the vertices of these contstraints if they were graphed? 30x+40y≤2100, 600x+900y≤45000, x≥0, y≥0 
And how would I graph them?

anonymous · Answer

graph each inequality individually...

for each, rearrange terms and solve for y so the inequality is of the form y ≤ mx + b  (or ≥ as appropriate).  Then, you can graph the inequalities just like you would graph a line.

There are four lines that form the edges of these inequalities, so you can find the vertices where the lines intersect.

anonymous · Answer

@oceanicmelody73  @Jessica_Moore   what do you think?  Does this sound reasonable as a way to work this problem?

anonymous · Answer

Yes, it does. Thank you. I'll try it.

anonymous · Answer

How do I make the inequalities into regular equations, with an equal sign instead of a ≤ or whatever?

anonymous · Answer

you don't really make them into regular equations, just think of them like that when you graph them.

graphing y > x is just like graphing the line y = x, but since it is an inequality, you would shade the area above the line to show that you are including all y values greater than x. 

(also, for greater than or less than, the line itself is a dashed line to indicate that the values ON the line are not included... when it's greater than or equal to, or less than or equal to, then use a solid line, since the line itself is part of the solution of the inequality.)

anonymous · Answer

Why didn't they explain it like that in class? haha thank you! So, if I act like the < is a = when graphing do I also do that while moving the equation around?

anonymous · Answer

That would make the first one y≤-.78x+52.5?

anonymous · Answer

close... your x coefficient is wrong... it would be -30x first, then divided by 40, it would become -0.75x or -3/4 x

anonymous · Answer

How do i graph x=0 and y=0?

anonymous · Answer

x = 0 is a vertical line that overlaps the y axis completely... it is ALL points where x=0, so not just at the origin but at all points above and below.

y = 0 is the same idea, but it's horizontal... y = 0 is the line of all points where y = 0, regardless of x... so it's the line that overlaps the x axis... all along the line, y = 0.

anonymous · Answer

Then what is the feasible region? I cannot tell from the online graphing calculator I'm using.