# Homework Help: Absolute valuing both sides probably a big no-no in Math

1. Sep 29, 2011

### flyingpig

1. The problem statement, all variables and given/known data
Suppose you are given the following inequality

$$-6x+ 4y+ 10z \leq 1203$$
$$7y + 15z \leq 1551$$
$$x,y,z \geq 0$$

Find a K > 0, if it exists, such that $$|x| \leq K$$, $$|y| \leq K$$, $$|z| \leq K$$

Flyingpig's take on this

So I am just going find the max of each x,y,z

Let x = 0, y = 0

$$10z \leq 1203 \iff z \leq 120.3$$
$$15z \leq 1551 \iff z \leq 103.4$$

So it makes sense to take $$z \leq 103.4$$ and that $$|z| \leq 103.4$$

Let x = 0, z = 0

$$4y \leq 1203 \iff y \leq 300.75$$
$$7y \leq 1551 \iff y \leq 221.57$$

Take $$y \leq 221.57$$ and that $$|y| \leq 221.57$$

Let y = 0, z = 0

$$-6x \leq 1203 \iff x \geq -200.5$$
$$0 \leq 1551$$

However $$x \geq 0$$, so $$|x| \geq 0$$

Side Question - not part of this thread

Was $$0 \leq 1551$$ useless? Also I know I am wrong, but I need to someone to give me a formal reason why I couldn't $$|x| \leq |-200.5|$$ and it becomes that $$|x| \leq 200.5$$

Side Question - ended, carry on with calculation below

So we have

$$|x| \geq 0$$
$$|y| \leq 221.57$$
$$|z| \leq 103.4$$

Now what does $$|x| \geq 0$$ imply? Does it say that any $$K \geq 221.57$$ will satisfy the inequalities $$|x| \leq K$$, $$|y| \leq K$$, $$|z| \leq K$$

2. Sep 30, 2011

### Staff: Mentor

I'm no expert here. But I wonder are you supposed to give this as the correct answer:

|x|,|y|,|z| <= 103.4

3. Sep 30, 2011

### cepheid

Staff Emeritus
Regarding the side question: I don't know if this qualifies as a "formal" reason why you can't do it, but it's pretty clear that you can't just take the absolute value of both sides of the inequality, because in general the inequality doesn't hold true after such an operation.

For example, if you have a > b where a = 3 and b = -4, then clearly it's not true that |a| > |b|, since 3 is not greater than 4.

4. Sep 30, 2011

### daveb

I would say the values for y and z are correct, but when the maximum possible values for y and z are plugged into the first equation, then x will need to be greater than (10z+4y-1203)/6

5. Sep 30, 2011

### flyingpig

Well no, x is unbounded. Isn't that the goal? Because if I go with your reasoning, then every unbounded set could be bounded by one K. K bounds the set/region

the word "set" and 'region" are the same here

6. Sep 30, 2011

### Staff: Mentor

I can't say what the goal is. But the question reads as "Find a K that satisfies these conditions ..." and it is unusual to reuse a symbol for different constants. I read "find a K" as singular.

Whatever was intended, I'd suggest the person who worded the question in this way and failed to see there may be an ambiguity should donate \$25 to a charity of his choice.

7. Sep 30, 2011

### flyingpig

Yeah I am sorry, this is part of another problem. I brought it into this context sorry. I originally had difficulty with solving for the inequalities