Macroeconomics Graph Problem *cleared* thanks

[fr]

http://pics.bbzzdd.com/users/frfr/ppf.jpg

Is that graph correct? It says the opportunity cost of 200 cars is 1,000 computers. Shouldn't point A meet at 800, 2000 instead of 700, 2000?

 

[simms]

No, because you're dealing with opportunity cost. *SEE NEXT POST*

As you increase the number of cars (from 700-900) lets say you have to decrease the number of computers you can make. Hence, with a rate of 1000 computers for every 200 cars, to increase your production by 1 car, you can't make 5 computers you originally had.

 

[simms]

If you're talking about the why A isn't at 800,2000 it's because the graph is showing a number of possibilities that are attainable by the maximium opp. cost. Whoops, now that I look at it... it should be 600, 2000 then 800, 1000 then 1000, 0. Does that make sense?

Sorry, it's been a while, but just trying to help.

 

[LordSnailz]

The graph is correct. Moving from point A to C, you get an additional 200 cars, (700 --> 900) but you lose 1000 computers (2000 --> 1000).

Dont' get confused with 1000 computer versus 1000 cars ... so you dont' need to start at 800.

 

[Wolverine27]

The graph is correct.

 

[Dudd]

It looks right to me. The labeling just isn't to scale. But to go from A to B, you would give up 1000 computers for 200 cars, or 1 to 5 like they say.

 

[sygyzy]

Graph is right. It's odd that the zero isnt at 500 though.

 

[ElFenix]

been a long time since algebra, eh?

 

[OulOat]

Originally posted by: sygyzy
Graph is right. It's odd that the zero isnt at 500 though.

No the graph is right, what they are saying in the bubble is the op. cost at that point.

 

[fr]

Thanks. I think I got it.

This is sad because I did well in macroeconomics a long time ago. Now this is review for other stuff. At least finals are half over now. :clock::wine: