Inequality problem

[Noproblem]

There is a simple inequality problem.
a and b are integers.
-4 < a < 5
-2 <= b < 6
what is the biggest value of 2a-b ?

---

I solved in this way.
-8 < 2a < 10
-6 <= -b < 2
and found this => -14 < 2a-b < 12
My answer is 11. But it is incorrect. The correct answer is 10.
But I wonder the reason. I think I solved this problem in proper way.

 

[Ishuda]

There is a simple inequality problem.
a and b are integers.
-4 < a < 5
-2 <= b < 6
what is the biggest value of 2a-b ?

---

I solved in this way.
-8 < 2a < 10
-6 <= -b < 2
and found this => -14 < 2a-b < 12
My answer is 11. But it is incorrect. The correct answer is 10.
But I wonder the reason. I think I solved this problem in proper way.

You are ok as far as you have gone but you must realize that if you reduce a by one [since it is a strict less than], you need to reduce the 12 by 2. That is if t=a - 1 [that is, a = t + 1] then
2*a -b = 2 (t+1) - b < 12
or
2 t - b <10
On the other hand, if you reduce b you can reduce the 12 by one. But how far can you reduce b?

The way I would do it is, to maximize 2a-b we want to make a as large as we can since 2a is adding to the total [think of it as +2a-b] and b as small as we can since it is being subtracted from the total. The largest a can be is 4 and the smallest b can be is -2 and 2*4-(-2) = 8+2 = 10.

 

[pka]

There is a simple inequality problem.
a and b are integers.
-4 < a < 5
-2 <= b < 6
what is the biggest value of 2a-b ?.

10 is correct. What is not correct is your effort to solve.

Look at \(\displaystyle -4<a<5\). The largest that \(\displaystyle 2a\) is ?

From \(\displaystyle -2\le b<6\), the largest \(\displaystyle -b\) is \(\displaystyle 2\). WHY?

What is the largest \(\displaystyle 2a+(-b)\) can be? Do not forget these are integers.

 

[Noproblem]

You are ok as far as you have gone but you must realize that if you reduce a by one [since it is a strict less than], you need to reduce the 12 by 2. That is if t=a - 1 [that is, a = t + 1] then
2*a -b = 2 (t+1) - b < 12
or
2 t - b <10
On the other hand, if you reduce b you can reduce the 12 by one. But how far can you reduce b?

The way I would do it is, to maximize 2a-b we want to make a as large as we can since 2a is adding to the total [think of it as +2a-b] and b as small as we can since it is being subtracted from the total. The largest a can be is 4 and the smallest b can be is -2 and 2*4-(-2) = 8+2 = 10.

Thank you very much sir. I understand it.

 

[Noproblem]

10 is correct. What is not correct is your effort to solve.

Look at \(\displaystyle -4<a<5\). The largest that \(\displaystyle 2a\) is ?

From \(\displaystyle -2\le b<6\), the largest \(\displaystyle -b\) is \(\displaystyle 2\). WHY?

What is the largest \(\displaystyle 2a+(-b)\) can be? Do not forget these are integers.

Thanks for the answer sir. I thought addition of two inequalities gives the answer. Because, I haven't seen any question like this. But, I realized that, I must look inequalities which are added, too.

 

[pka]

Thanks for the answer sir. I thought addition of two inequalities gives the answer. Because, I haven't seen any question like this. But, I realized that, I must look inequalities which are added, too.

The point is that these are thought exercises. They are not meant to have anything done to them.

Given \(\displaystyle -5<x\le 4~\&~-3\le y<6\) (these are integers). What is the minimum \(\displaystyle -3x+2y~?\).

Now the answer \(\displaystyle -18\). HOW?