Exam 1 Graphs & Relations Answers

[hifer]

My answers..
1. D
2. C
3. E
4. B
5. D
6. E
7. C
8. B
9. A

[reg]

Graphs and relations Q9.

Objective function: Z = 2x + ky, maximum occurs only at (100, 0).

Hence

200 = 2x + ky
y = (200 - 2x) / k

Function exists at both (100, 0) and (50, 50), this can only be true when k = B = 2 (not k = 1, A).

Unless I'm horribly wrong .

[SilverBullet]

got 4 or five right which is not a bad result for myself

[Odette]

Graphs and relations Q9.

Objective function: Z = 2x + ky, maximum occurs only at (100, 0).

Hence

200 = 2x + ky
y = (200 - 2x) / k

Function exists at both (100, 0) and (50, 50), this can only be true when k = B = 2 (not k = 1, A).

Unless I'm horribly wrong .

Lol I got B for Qn 9 as well ... worked it out differently though

[hifer]

Graphs and relations Q9.

Objective function: Z = 2x + ky, maximum occurs only at (100, 0).

Hence

200 = 2x + ky
y = (200 - 2x) / k

Function exists at both (100, 0) and (50, 50), this can only be true when k = B = 2 (not k = 1, A).

Unless I'm horribly wrong .

umm, if K = 2, at  (50,50) Z also = 200

Which means there 2 maximums, which can't be true...

[reg]

Haha maybe, I was under the impression the function had to go through (50, 50). Apparently not. Somebody else input plz :p

on second though you are right.

scratch that, i have no idea anymore. thanks :@

[Defiler]

Same except for 7, which I got B.

[Odette]

Damn, I originally had A for Qn9, why didn't I leave that answer!!! grr

[gulamali]

Same except for 7, which I got B.

how did you get B ... i got E for this question, knowing me I'm probably wrong ;[

[Defiler]

Okay, what I did was...

y = x^3

y = 1, x = 3

hence

1 = 3^3

therefore

y=(1/27)x^3

if we are going to x...

y=cubert(1/27)x^3

y=(1/3)x

sub in (3,1)

1=1

I would of picked E if I assumed (1/27) was a constant.

But it appears that I've done it wrong.

[Odette]

Same except for 7, which I got B.

how did you get B ... i got E for this question, knowing me I'm probably wrong ;[

I got E too...but it's wrong i think

[Nick]

Same except for 7, which I got B.

how did you get B ... i got E for this question, knowing me I'm probably wrong ;[

I got E too...but it's wrong i think

I got E aswell.

[dxb]

Same except for 7, which I got B.

how did you get B ... i got E for this question, knowing me I'm probably wrong ;[

I got E too...but it's wrong i think

I got E aswell.

Got E too

I'm pretty sure its wrong, because the graph represents a x3 transformation already, so to undo this, you would have to square root it and not cube it again.

Meh, I've got no idea.

[Odette]

Wow the 4 of us got E... hmm

[Nick]

Can someone explain to me how it can't be E?