Question

Help on linear inequalities please!
If x,y,z≥0, 3y+2z=3+x, 3y+z=4-3x, find the largest and smallest value of 3x-3y+4z. Every time I've done this I get a different answer. Thank yoU!

Answer 1

do you know linear algebra?
you can write the augmented matrix
[ -1 3 2 3; 3 3 1 4]
and find its reduced row echelon form
[1 0 -¼ ¼; 0 1 7/12 13/12]

which means
x= ¼ z + ¼
y= -7/12 z +13/12

and 3x-3y+4z can be put in terms of z

Answer 2

if I did the math correctly, we get
3x-3y+4z= (26/4)z -10/4
for z>=0 that means the smallest value is -10/4 and the largest is infinity

Answer 3

:)

Answer 4

:( thank you but the answer is actually -5/2≤3x-3y+4z≤67/7 which means it cant be infinity right? oh wait can it?

Answer 5

@phi i got the same answer as you as in x= ¼ z + ¼
y= -7/12 z +13/12
After which I put
1/4 z + 1/4 ≥0
z+1≥0
z≥-1

and

-7/12 z + 13/12 ≥0
-7z+13≥0
z≥13/7

This doesn't make much sense to me actually since hten the answer would be z≥13/7. And this is only the lower limit, and i don't know how to find the upper limit

Answer 6

make one equation in one variable

Answer 7

If x,y,z≥0, 3y+2z=3+x, 3y+z=4-3x, find the largest and smallest value of 3x-3y+4z...\[x=3y+2z-3.(1)....x=(1/3)(4-3y-z)......from these two equations \]gety and z in terms of x and subistitute with them to in the form of equation in one variable

Answer 8

I know that's what I did. I made everything in terms of z. Afterwards, I used the z form of it to write: x≥0 (x in z form) and y≥0 (y in z form e.g. 2z) and solved it. But I wasn't able to get a range from this, I only managed to get one value. E.g. z≥2. This isn't a range and hence I'm unable to solve the question

Answer 9

oh, you are correct!

I just noticed that when we say
x= ¼ z + ¼
y= -7/12 z +13/12

we also have to remember x>=0. The first equation is valid for all z>=0
but in the 2nd equation
we need
\[ y\ge0\\ -\frac{7}{12} z +\frac{13}{12} \ge 0 \\
\frac{13}{12} \ge \frac{7}{12} z \\
z \le \frac{13}{7}
\]

so we have an upper bound on z (else y would be negative)
using that upper bound for z (not infinity!) we get
\[ 3x-3y+4z= \frac{13}{2} z -\frac{5}{2} \\= \frac{13}{2}\cdot \frac{13}{7}-\frac{5}{2}\\
= \frac{67}{7}
\]
as the upper bound

Answer 10

so the final result is
\[ -\frac{5}{2} \le 3x-3y+4z\le \frac{67}{7} \]

Answer 11

***
-7/12 z + 13/12 ≥0
-7z+13≥0
z≥13/7
****

I think that is where you ran into trouble. you have to "switch" the relation operator > to < if you multiply or divide by a negative number.
However, it is always safe to add/subtract values from both sides.

For example, you could start with
-7z+13≥0
and add +7z to both sides to get
13 ≥ 7z
13/7 ≥ z
which is the proper relation.

Answer 12

and just to complete this, when you found
***
1/4 z + 1/4 ≥0
z+1≥0
z≥-1
****

we are told z ≥0, so we have to use 0 as at the lower bound on z, not -1

Answer 13

THANK YOU!! :D