what point in the feasible region maximizes the objective function?

Question

tester97 · Answer

attachment

anonymous · Answer

ok we should sketch this region first.. can you to it ?

tester97 · Answer

can i wat?

anonymous · Answer

can you sketch the region for me?

tester97 · Answer

how?

anonymous · Answer

ok look at y <=-x+3
now first we sketch y=-x+3

anonymous · Answer

the attchment wont work imma have too draw it out hold on it take a bit

tester97 · Answer

hold on im goin to try to graph it

anonymous · Answer

okay

anonymous · Answer

|dw:1382984800293:dw|
look at y <=-x+3
now first we sketch y=-x+3

tester97 · Answer

yea mine looks something like that on the graphing calculator

anonymous · Answer

this line is y = -x +3 agree ?

tester97 · Answer

$$y \le-x+3$$

anonymous · Answer

yes now we will add the graph y = (1/3)x + 1

anonymous · Answer

|dw:1382985016406:dw|
Like this. Do you have this?

tester97 · Answer

my graph is scrambled up so idk...........

anonymous · Answer

well something like that since the intersection of  y =(1/3)x + 1 and y = -x +3
is 
-x  +3 = (1/3)x + 1
-(4/3)x =-2 
x = 1.5

anonymous · Answer

Are you understanding this?

tester97 · Answer

noooo im not using equal signs im using inequality signs so this it totally different

anonymous · Answer

it has the same value I just put an equal sign sorry if I confused you. I'm trying here math is hard for me too...

anonymous · Answer

I probally should of mentioned this it would of helped
so now the region that satisfies 
x >= 0
y >=0 
is the first quadrant right ?

anonymous · Answer

Since the first quadrant has only positive x values and positive y values

tester97 · Answer

ahhhhh nvm i got the answer now it was so obvious *facepalm* thanks for your help thought

anonymous · Answer

So the awnswer would be (3,0)

anonymous · Answer

Glad I could help you though. Good luck :)