D1 & D2 June 2014

I would kill to retake my AS level choices, I should have done Economics and History, I loved History at GCSE
Hahahahahahah okay I won't take you seriously :P

I'm on Edexcel, which seems pretty easy going in comparison to the other boards this year ^_^

Ooh I wanted to go to Warwick but now I'm not too sure because I think at worst I might get ABBC and ruin my chances of pretty much every university for my course xD

Same but at a-level it gets tough, I dropped history for A2... the exam is so risky

Edexcel as well indeed!

Well you never know, whats your course btw? I'd still apply to Warwick imo, no reason not to. My insurance is Durham, would rather not go there since its on the edge of England about 5hrs from London xD

is there any point doing the old spec D1 exam papers from before 2009? How much content have they taken out/Added?

2/10 is how ready I am for it, but I have almost 2 weeks space between my D1 and D2 exams and the other maths modules are far more important :P

Nice to know I'm making my mark on TSR as the crazy goat-lover/worshipper

JOIN US :3

Good luck for that.

Nah, I'm a goat-eater not a goat-worshiper.

;((((((((( that's so sad

Don't declare war. We've shoot them all

*weeps into the distance*

Q7d) linear programming - how do you find the minimum total number of boxes when the objective function is given in question part (e)?
i've seen the mark scheme but i don't get it

If you have drawn R correctly, all you need to do is use the vertex testing method (which if you don't know how to do, is in pg 135 of the Edexcel D1 textbook). You look at all the corners/vertices of the feasible region R and locate the (x,y) values of all these points. Then you simply add them (the x and y coordinates) together as x + y is the total number of boxes , and the one which gives you the minimum total is the minimum total number of boxes he needs to prepare each day. This is the vertex (10,60) in R, and 60 + 10 = 70 which is the answer they want. You do not need a profit line for it.

what if you used the constraint given in (a) x + 2y < 160? would that still be correct?

How would you do that way?

You could do as scientific222 said or use a profit line/objective line.

If you want to use an objective line way this line would be different to that in (e), because for this we want the minimum number of total boxes (and not what they want in the later parts).

The total number of boxes is x + y....so the objective function for this would be Minimize P = x + y

So you could draw a profit/objective line. This would be crossing the x and y axes at the same value. So you use a ruler to drag this line and find the vertex to which it touches first. (Because we want a minimum)

If you don't like this way you could do as scientific222 has said

i understand the method, it's just that i got confused with (d) and (f). initially, i didn't get the difference but now i think i do

so just to confirm, the answer to (d) is not neccesarily the optimal solution? all they're asking for is the minimum possible total number of boxes if the guy is to make a profit?

however (f) is where we find the optimal solution, so (f) has nothing to do with the answer found in (e)?

Yes the minimum possible number if all constraints are to be considered.

Most certainly no

Quite confused with this transportation problem... (from Question 3 of June 10).

Would anyone be able to explain why the 0 is placed in the specific places shown in the mark scheme to resolve the degeneracy? I was under the assumption that the 0 could go in any of the unused cells, but here they've placed them in different places depending on the exiting cell...?

Mark scheme:

No you could only place the 0 in on of the two possible exiting cells. The textbook makes this clear, doesn't it?

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My textbook just says "the cell must be chosen so that it is possible to calculate all the shadow costs....it is usual to choose the cell with the lowest shipping cost". But then why don't you put the 0 in BY?