|
Math Problem
|
View this Thread in Original format
| creon444 |
Can anyone of you help me out with this problem. I've
been trying to solve it for ages and still don't
know where to even start. It seems nearly impossible to
solve. So I'm asking if there are any smart people here
that are willing to help? ANY hints/ideas would be
greatly appreciated! :)
Suppose you need to buy some blue and red pens. The
store you go to sells red pens for 17 cents apiece and
blue pens for 13 cents apiece. You only have $4.95 to
spend. You need to buy as many pens as possible, but
you also want to buy as few red pens as possible.
Given that the difference between the number of blue
pens and the number of red pens cannot be more than 5,
how many blue and red pens can you buy in total? |
|
|
| You aint Ninja |
| One million. |
|
|
| You aint Ninja |
-edit-
double post :nervous: |
|
|
| butterfly |
i can give you the answer buit then you wouldnt be doing your homework.
they gave you enough information to solve a 2 variable system of algebraic equations.
you know that however many you buy, this will be true:
(.17)R + (.13)B < 4.95
you also know that the difference wont be more than 5 AND you will have more blue than red, so:
B-R < 5
so solve this system of equations.
edit: i ed up the first time cause i cant read but i fixed it. |
|
|
| creon444 |
| quote: | Originally posted by butterfly
i can give you the answer buit then you wouldnt be doing your homework.
they gave you enough information to solve a 2 variable system of algebraic equations.
you know that however many you buy, this will be true:
(.17)R + (.13)B < 4.95
you also know that the difference wont be more than 5 AND you will have more blue than red, so:
B-R < 5
so solve this system of equations.
edit: i ed up the first time cause i cant read but i fixed it. |
But what about that thing where they say you need to buy as few red pens as possible. Besides, those are not equations, but inequalities. I know how to solve a system of equations, but this is not the case!
Thanks anyway :) |
|
|
| butterfly |
| quote: | Originally posted by creon444
But what about that thing where they say you need to buy as few red pens as possible. Besides, those are not equations, but inequalities. I know how to solve a system of equations, but this is not the case!
Thanks anyway :) |
you used that information. they told you both that you want fewere red pens but that the difference cant be more than 5. so that is how you got the equation B-R < 5
anyway, you solve it just like the inequalities are equal signs. the only time you worry about it is if you are multiplying by a negative.
so this is how it solves:
.17R + .13(5+R) < 4.95
.17R + (.13)(5) +.14R < 4.95
.3R < 4.95-(.13)(5)
r < 14.3
b < 5+14.3
so you know you cant have a partial pen, so R must be 14, which makes b 19.
when you go back and check, you see that your total money spent is $4.85, which doesnt leave you any money to buy any more pens.
(and check these numbers. i always up my calcuations but the logic here is right.) |
|
|
| kaige |
14 red pens @ 17 cents = $2.38
19 blue pens @ 13 cents = $2.47
33 pens @ $4.85 |
|
|
| Fundamental |
| quote: | Originally posted by butterfly
so you know you cant have a partial pen, so R must be 14, which makes b 19.
when you go back and check, you see that your total money spent is $4.85, which doesnt leave you any money to buy any more pens.
(and check these numbers. i always up my calcuations but the logic here is right.) |
Moral of the story is...
YOU CANNOT HAVE RED PEN15!! |
|
|
| You aint Ninja |
| join the pen15 club. |
|
|
| Acid Junkie |
| quote: | Originally posted by butterfly
i can give you the answer buit then you wouldnt be doing your homework.
they gave you enough information to solve a 2 variable system of algebraic equations.
you know that however many you buy, this will be true:
(.17)R + (.13)B < 4.95
you also know that the difference wont be more than 5 AND you will have more blue than red, so:
B-R < 5
so solve this system of equations.
edit: i ed up the first time cause i cant read but i fixed it. |
that is too easy a solution u've got there. give the man something he can show off with :haha:
dude, if u want to get an A++ in ur class, show them sometingh like this. this is how to solve the problem using 3(!) variables.
let x be the number of red pens you can buy and y the number of blue pens. then, given that u can only spend $4.95=495 cents, we have
17x+13y≤495
x≥0, y≥0
since the difference of the number of both kinds of pens cannot exceed 5, we have
|x-y|≤5
we need to find the biggest value that x+y=z can take
so what we have here is a nice mixed system with 3 variables
17x + 13y≤495
|x-y|≤5
x+y=z
0≤x≤z
0≤y≤z
applying a few substitutions for y (y=z-x), we get
4x+13z≤495
|2x-z|≤5
which is the same as
x≤(495-13z)/4
(z-5)/2≤x≤(z+5)/2
considering that 0≤x≤z, we can rewrite the above system as
z≥0
(z+5)/2≥0
0≤(495-13z)/4 <=> 495-13z≥0 <=> z≤495/13
(z-5)/2≤(495-13z)/4 <=> 2(z-5)≤(495-13z) <=> 15z≤505 <=> z≤505/15
or as
0≤z
-5≤z
z≤495/13
z≤505/15
or as 0≤z≤505/15=101/3=33+2/3
which means the greatest integer z can be is 33
so we need to find the smallest x possible when z=33. using 33 for z, we get another system with only one variable
x≤16.5
14≤x≤19
0≤x≤33
the smallest solution of which is 14.
same answer :D |
|
|
| whiskers |
^^^^^^^^^^^^^^^^^^^^^^^
that is weak
use linear algebra and double integrals and write a 20-pages detailed solution.
that'll surely make your teacher give you an A+ and she'll treat you like her own child for the rest of the year |
|
|
|
|
