Question

A photography shop owner displays two sizes of photo in his shop. Each photo A requires 720cm of display space and each Photo B requires 360cm of display space. The available display space cannot exceed 108,000cm .

The shop owner buys each photo A for €6 and each photo B for €8. He does not need to spend more than €1200.

(i) Taking x as the number of photo A and y as the number of photo B, write down two inequalities in x and y and illustrate these on graph paper. (8 marks)

Answer 1

(ii) During a promotion the selling price of photo A is €12 and of photo B is €10. Assuming that the shop owner can

Answer 2

I think (i) asks for the constraints?

Answer 3

ending of the second question that got cropped out ", How many of each type should he display in order to maximise his income? (4 marks)"

Answer 4

And I'm not quite sure still what finding the constraints means

Answer 5

What limits your linear program

Answer 6

So, the available display space cannot exceed 108000... so...

Answer 7

So 720x + 360y = 108,000

Answer 8

Yes, actually.

Answer 9

Or simplified as 2x + y < 300

Answer 10

No, wait...
more like
\[\huge 720x+360y \le108000\\ \\ \huge 2x+y\le300\]

Answer 11

As for the other one..."He does not need to spend more than 1200"

Answer 12

I'm still clueless as to how I'm going to finish this and write it on a graph, let alone do the second one

Answer 13

Just graph them as lines first...
2x + y = 300
and the other one, which I'm certain you're already got.

Answer 14

Pretty sure I'm suffering from my usual dose of the overcomplicating the whole thing, the 2x means that the slope will move down 1 x for every 2 "Y"s right?

Answer 15

Yeah, seems like it :)

Answer 16

Still haven't figured out the second one :p

Answer 17

Well, that's the problem, isn't it? :D
Now, If you have graphed your two inequalities, at which point did they intersect?
^(honest question, I don't know, myself, so you have to tell me :) )

Answer 18

I'm currently mult-tasking my studying by also doing quadratic functions at the same time so I haven't graphed this just quite yet xD

Answer 19

take your time :)

Answer 20

Just so we're clear, I pick an arbitrary point between 0 - 150 and graph it like that, right?

Answer 21

Something like that?

Answer 22

Arbitrary points? I don't really know how you were taught to graph lines, but just graph them and solve for their point of intersection... and yeah, I guess it looks something like that...

Answer 23

Mainly self taught with little formal education in between.

Answer 24

Okay, so where do they intersect?

Answer 25

I don't know, I am ill-equipped with the knowledge of how to solve this.

Answer 26

Okay... at least give me the two lines that we're looking at (you've already given one of them)

2x + y = 300

what's the other one?

Answer 27

3x + 4y = 600

Answer 28

Okay, we have
2x + y = 300
3x + 4y = 600

Let's look at the first equation...
2x + y = 300
Try to solve for y (in terms of x)

Answer 29

y = - 2x + 300

Answer 30

?

Answer 31

Sorry...
Now, using this value for y, replace y in the OTHER equation with this expression in terms of x.

Answer 32

y = - 0.75x + 150

Answer 33

No... In the first one, you got
\[y =\color{red}{ -2x-300}\]
what I want you to do is substitute this expression for y
for the y in the OTHER equation.
\[3x + 4\color{red}y=600\]

Answer 34

So 3x + 4(-2x - 300) = 600?

Answer 35

Yes. and now, solve for x.

Answer 36

3x - 8x - 1200 = 600
-5x = - 600
x = - 120?

Answer 37

= 600*

Answer 38

So.... x = ?

Answer 39

X = - 360

Answer 40

I suggest you do this again, from the beginning...
and pay close attention to each step.
3x + 4(-2x - 300) = 600

Answer 41

Wait a minute, I made a typo O.o
Anyway...
\[\Large y = -2x \color{red}+300\]

Answer 42

Ah, so I was right when I made a blunder :p

Answer 43

It tends to happen.
Now, let's do this more seriously, and without error, this time.
Substitute, and solve for x, you get?

Answer 44

3x - 8x + 1200 = 600
3x - 8x = 600 - 1200
-5x = -600
x = 120

Answer 45

Okay, good.
Now substitute this value for x into any one of the two equations you started with, and solve for y.

Answer 46

2(120) + y = 300
240 + y = 300
y = 60

Answer 47

Good. Now, in the case of these maximization problems, the optimal (maximum) solution would lie in one of the corners of your enclosed region.
Obviously, (0 , 0) would be a corner of your enclosed region, but it isn't the maximum value.

What are the other 3 corners?

Answer 48

(0 , 60)
(120, 0)
(120, 60) ?

Answer 49

(0,60)
and
(120,0) are wrong... but you got the right idea... zero out one of the variables and max out the other...

Remember your two inequalities?
\[2x+y \le 300\\ 3x+4y\le 600\]

Answer 50

Now, if we let x be zero, what results in the two inequalities for y?

Answer 51

y <= 300
y <= 150

Answer 52

Good. Now, pick the "less picky" of these two inequalities.
What I mean is pick the one which, if satisfied, would also satisfy the other inequality.

Answer 53

300

Answer 54

...is a number, not an inequality.

Answer 55

y <= 300

Answer 56

Okay... is it true that if y <= 300
then automatically, y <=150 ?

Answer 57

Well 150 is less than 300 so it would 150 would be an inequality within 300.

Answer 58

Or am I viewing this completely wrong.

Answer 59

Yes or no.
If y <= 300
does that already imply that
y <= 150 ?

Answer 60

Yes.

Answer 61

Really?
What about 200?
200 <= 300
but
200 <= 150 ?

Answer 62

Ah right, I was thinking more in terms of what would cover the most of the inequality.

Answer 63

No. You pick the inequality that would satisfy the other.

Answer 64

So, one of the corners, since we let x = 0
is
(0 , 150)

Now find the other one, by letting y = 0 instead.

Answer 65

\[2x+y \le 300\\ 3x+4y\le 600\]

Answer 66

150 < 300
200 < 300

Answer 67

Blergh

Answer 68

x < 150
x < 200

Answer 69

Okay, so which one of these do you choose?

Answer 70

x < 150 as it satisfies both itself and x < 200

Answer 71

Good. So your other corner point is...?

Answer 72

(0,0) ( 120, 60) ( 0, 150) (150 , 0)

Answer 73

Good. Now remember the function we were trying to maximize?
The one where picture A sells for 12 and picture B sells for 10?
Assuming all can be sold, what is the profit function
\[\Large f(x,y) = ?\]

Answer 74

\[f(12, 10) = ?\]

Answer 75

What?

Answer 76

If the seller sells x units of picture A
and y units of picture B
How much does the seller make?

Answer 77

x + y = z

Answer 78

12x + 10y > 1200?

Answer 79

I'm not asking for an inequality, I'm asking for the function...
f(x,y) = ?

Answer 80

The function that which we are trying to maximize. (the profit function)

Answer 81

Stuck?
f(x,y) = 12x + 10y

Answer 82

Ah, I found myself wrapped in economic research notes looking for tthe profit function.

Answer 83

Don't overcomplicate things... this isn't a profit function, strictly speaking, in terms of economics, this is actually a revenue function...
ANYWAY

Now, all you have to do is test out your four corner points, see which one gives you the biggest value.

Answer 84

...for f(x,y)

Answer 85

I gotta rush, but Ive got it, I've rached the part of the textbook that looks simple.