VCE General & Further Maths Question Thread!

[kavmeh]

I didn't even realise there was a table there   Should read the questions carefully
Thank You so much

[n.a]

Hi guys,

Can someone help me with VCAA 2007 Exam 1 Question 9 from Graphs and Relations?

Thanks!

[Lawyer]

Hi guys,

Can someone help me with VCAA 2007 Exam 1 Question 9 from Graphs and Relations?

Thanks!

Okay so the question asks you for to maximise at y= 2x + ky. And it has to be at the point (100,0)

Immediately from this question i get the feeling that the lower k the better. Why? Because the point is asking at (100,0) So it does not matter what k is, because k times y= 0. However the greater the value K is then the greater all the other points are.

You have 5 choices: A) 1 B) 2 ect. If K=1, then it would maximise (100,0) because all the other points (50,0) (25,25) and (50,50) would maximise the objective function as k increase.

So to maximise the point (100,0) at 2x+ky then k would have to be the smallest, so that (100,0) = 200, would not be smaller than the other points.

Further explanation: Lets take k=1. Then 2x100+1x0=200. If we substitute 1 into the other points, the max value would still be 200, which is at (100,0)
However if we take 2. The 2x100+2x0= still equals 200, but it is not the only maximum value as now the points (50,50) is. So 2x50+2x50=200. So if it was 2 then the maximun value would exist at (100,0) and (50,50) but the question asks for only at (100,0) so it has to be lower than 2, which is 1, thus it is A

[MightyBeh]

Hi guys,

Can someone help me with VCAA 2007 Exam 1 Question 9 from Graphs and Relations?

Thanks!

Probably not the fastest way to do it, but I'd first cross out 4 and 5 (because they're silly, try checking the function with those at (50,50)) then define a function so I could quickly test the rest:

Code: [Select]

Define Z(k,x,y)=2x+(k*y)
Then you can test each one using the following

Code: [Select]

z(k,x1,y1) ≥ z(k,x2,y2)
So testing k=1;

Code: [Select]

z(1,50,50) ≥ z(1,100,0)comes up with false, so we know that the point (100,0) is larger than (50,50) when k = 1.

Making sure that this is consistent, testing k = 2 comes up with 'true' (because they're equal at this point, Z=200).

Testing k = 3 in the same format also comes up true so that can't be our answer.

I picked (50,50) as the testing point because the others seemed kinda far out but in exam conditions I'd probably recommend checking them at least once. So the answer is A. 1

Edit: Damn, Lawyer beat me to it 

[n.a]

Thank you so much, Lawyer and MightyBeh!

[kavmeh]

Hi,
Can anyone please help me with question 9 from 2008 exam 1 (Graphs and Relations) Why isn't the answer E?
and also questions 8 2009 exam 1 (Graphs and Relations)
Thank you

[n.a]

Okay, I can help you with the 2009 Question 8:

-So we know that he has a fixed cost of $150 and he also needs to pay each worker $20 which is also part of his cost.
-There are 3 workers, so the actual cost is $60 per hour, if you take into account all the workers he needs to pay.
-Therefore, your cost equation is C=60x+150, where x is the number of hours worked, and C is cost.
-The question tells us that the job takes 4 hours, so now sub in x=4 into the cost equation and you get Cost= $390.
-Now for the revenue cost, we don't the amount he charges, but we do know that it takes 4 hours, and he charges by the hour.
-Therefore, your revenue equation is: R=4p, where R is revenue, and p is the price he charges.
-To break even, the cost should equal revenue so: 390=4p.
-Solve for p. 
-Ta-dah! Hope that helps!

[kavmeh]

Thank you so much n.a
You made it so much easier to understand the question

[n.a]

No worries, and I'm sorry I couldn't help with the other one, only I got that wrong too, so...can't help much. But I'm sure the amazing AN people will be on it soon!

[bae]

Hi,

I'm really stuck on question 2c and onwards in matrices on last years exam 2, could anyone explain to me how I'm supposed to keep the state matrix constant?

Thanks 

PS, I got 36/40 on exam 1 and 47/60 on exam 2 (lower than usual) anyone know what kind of SS I'm in for??

[paper-back]

Hi,
Can anyone please help me with question 9 from 2008 exam 1 (Graphs and Relations) Why isn't the answer E?
and also questions 8 2009 exam 1 (Graphs and Relations)
Thank you

For Question 9;
We know that Profit = Revenue - cost
We can determine the cost function through finding the equation of the straight line which is:
C = 2x + 50
Since the person has sold 150 cards, their cost will total to: $350 (2x150+50=350)
The person has sold 150 cards, so their revenue will be 150 x y (where y = selling price)
Since the profit is $175, we can plug all these information into the P= R -C equation so that;
175 = 150y+350

Solve for y, and you get the selling price as $3.50

EDIT: holi moli, ignore me. I thought you were asking for question 9 of 2009

[MightyBeh]

Hi,
Can anyone please help me with question 9 from 2008 exam 1 (Graphs and Relations) Why isn't the answer E?

E is correct if you forget (like I did the first time I did the question ) that only the corner points of the feasible region are to be included in your calculation (I'm looking at you, (0,100)  ).

To (accurately) find the coordinates of point M we first need to know the equation of the lines and solve for the intersection. I'm too lazy to type it out so I'm going to assume you had no issues with this part and know that the coordinates of M are (20,50). (Let me know if you did have issues though and I'll show you <3 )

So our corner points are: (0,60), (20,50), (40,0) and (0,0). Obviously we don't need to include 0,0 though because none of the equations are going to be maximised there.

Then I just subbed in the three reasonable points to each equation (thankfully the answer is A.)

(0,60)
(M)
(40,0)

So the function Z = x + y is maximised at the point M.

But I'm sure the amazing AN people will be on it soon!

[paper-back]

Hi,

I'm really stuck on question 2c and onwards in matrices on last years exam 2, could anyone explain to me how I'm supposed to keep the state matrix constant?

Thanks 

PS, I got 36/40 on exam 1 and 47/60 on exam 2 (lower than usual) anyone know what kind of SS I'm in for??

Are you referring to where it states that "the number of voters in the city is expected to remain constant until the election is held in June?"

[bae]

Are you referring to where it states that "the number of voters in the city is expected to remain constant until the election is held in June?"

Yes exactly that, I don't get how to keep it constant

[MightyBeh]

Hi,

I'm really stuck on question 2c and onwards in matrices on last years exam 2, could anyone explain to me how I'm supposed to keep the state matrix constant?

Thanks 

PS, I got 36/40 on exam 1 and 47/60 on exam 2 (lower than usual) anyone know what kind of SS I'm in for??

What's your SAC average?

Workings for 2c - 2d are attached, did you need help with question 3 as well? 

Edit: Do you mean in the part before 2b where it says 'the number of voters in the city is expected to remain constant'? 90% sure that just means the sum of all the values in any state matrix will add up to 12000.