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SonyGmat
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Inequalities using extreme values. 15 Nov 2011, 05:20
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I have a question regarding inequalities using extreme values.

1st Example

Lets say that I know that \(y>4\) and I want to find the values that x can take. Moreover, I have 2 more inequalities:

\(x-y>-2\)

\(x-2*y2\).

How can we solve it using extreme values?

I know that y must be between 4.00000001 and a value like 10000000000.

If I use the smallest possible value to the first equation: x must be a bit bigger than 2
If I use the largest possible value to the first equation: x must be a bit bigger than a number really really big.

How do we continue our reasoning??

2nd Example.

Lets say that we have 2 inequalities \(a^2*b>1\) and \(b<2\) and I want to find the values that a can take.

From this we can understand that b must be between 0 and 2 because if b was negative then the first equation would be wrong. (I can't figure out how to solve this algebraically and I would really like to see how, if possible).

Attempt using extreme values

Assume that b is a bit bigger than 0, lets say 1/10000000 we can conclude that \(a^2\) will be bigger than a very big number. Squaring both sides we can see that a is larger that a big value and smaller than a very small.

How do we continue our reasoning...



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Inequalities using extreme values. 15 Nov 2011, 09:20
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SonyGmat

Lets say that we have 2 inequalities \(a^2*b>1\) and \(b<2\) and I want to find the values that a can take.

From this we can understand that b must be between 0 and 2 because if b was negative then the first equation would be wrong. (I can't figure out how to solve this algebraically and I would really like to see how, if possible).
Show more

Okey I found out how I can solve it algebraically.

\(b*a^2>1\) which equals \(b > \frac{1}{a^{2}}\) (we can divide by \(a^2\) since it is always positive)

Using the inequality we just found with \(b < 2\) we conclude that \(\frac{1}{a^2} <b <2\). Therefore \(a^2 > \frac{1}{2}\)

Anyone can show me how to solve it using extreme values?
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Inequalities using extreme values. 15 Nov 2011, 13:42
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SonyGmat
SonyGmat

Lets say that we have 2 inequalities \(a^2*b>1\) and \(b<2\) and I want to find the values that a can take.

From this we can understand that b must be between 0 and 2 because if b was negative then the first equation would be wrong. (I can't figure out how to solve this algebraically and I would really like to see how, if possible).

Okey I found out how I can solve it algebraically.

\(b*a^2>1\) which equals \(b > \frac{1}{a^{2}}\) (we can divide by \(a^2\) since it is always positive)

Using the inequality we just found with \(b < 2\) we conclude that \(\frac{1}{a^2} <b <2\). Therefore \(a^2 > \frac{1}{2}\)

Anyone can show me how to solve it using extreme values?
Show more

Once you have: \(\frac{1}{a^2} <b <2\)

Then you can test various values including extreme ones to find the possible values for a.

If a= something super big
then, \(\frac{1}{a^2}\) essentially goes to zero and you get:
0 < b < 2 (good)

If a = 1, then you get 1 < b < 2 (good)
If a = 1/2, then you get 4< b < 2 (no good)
If a = 1/4, then you get 16 < b < 2 ( no good)

So actually looks like the pivot point is when a = 1/(root 2), and you would get 1 / (a^2) = 1 / (1/2) = 2
Plug it in and you get: 2 < b < 2 (which is right at the pivot point).

Anything less than a = 1 / (root 2) is no good.
Anything more than a = 1/ (root 2) is ok.

So a can be > 1 / (root 2) which is the same as saying \(a^2 > \frac{1}{2}\)
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Inequalities using extreme values. 15 Nov 2011, 14:21
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Once you have: \(\frac{1}{a^2} <b <2\)

Then you can test various values including extreme ones to find the possible values for \(a\).
Show more

if you find this, you can simply multiply by \(a^2\) and divide by 2 and find the range of \(a\).

The question is when you can't figure out this relationship between \(a\) and \(b\)and you just know that you have those 2 inequalities \(a^2*b>1\) and \(b<2\), how can you find the values that \(a\) will take using extreme values?

For example we know that \(b\)will be a bit less that 2 and a bit more than 0... then what?
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Inequalities using extreme values. 21 Nov 2011, 07:40
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SonyGmat
I have a question regarding inequalities using extreme values.

1st Example

Lets say that I know that \(y>4\) and I want to find the values that x can take. Moreover, I have 2 more inequalities:

\(x-y>-2\)

\(x-2*y<-6\)

I know that by multiplying the first one by -2 (which changes the sign) and then adding the two aforementioned inequalities I can find that \(x>2\).

How can we solve it using extreme values?

I know that y must be between 4.00000001 and a value like 10000000000.

If I use the smallest possible value to the first equation: x must be a bit bigger than 2
If I use the largest possible value to the first equation: x must be a bit bigger than a number really really big.

How do we continue our reasoning??

2nd Example.

Lets say that we have 2 inequalities \(a^2*b>1\) and \(b<2\) and I want to find the values that a can take.

From this we can understand that b must be between 0 and 2 because if b was negative then the first equation would be wrong. (I can't figure out how to solve this algebraically and I would really like to see how, if possible).

Attempt using extreme values

Assume that b is a bit bigger than 0, lets say 1/10000000 we can conclude that \(a^2\) will be bigger than a very big number. Squaring both sides we can see that a is larger that a big value and smaller than a very small.

How do we continue our reasoning...
Show more

I dislike 'making equations and solving' but when it comes to solving some inequalities simultaneously, I generally prefer to tame my imagination and follow algebra because it is risk free. Let me discuss how to use algebra in both cases and then I will talk about 'guessing the range'

Golden Rule: Given two inequalities, you can always add them up if the inequality signs point in the same direction.

\(x-y>-2\)
\(x-2*y<-6\)

Here, the signs point in opposite directions.

To get the range of x, we multiply the first inequality by -2 and add.
\(-2x+2y<4\)
\(x-2y<-6\)
-x < -2 or x > 2

To get the range of y, we multiply the first inequality by -1 and add.
\(-x+y<2\)
\(x-2y<-6\)
-y < -4 or y > 4

We get the range of x and y.

\(a^2*b>1\)
\(b<2\)

To get the range of a, multiply the second inequality by -a^2 (which we know will be negative since a^2 is positive) and add. (just solving them blindly)
\(a^2*b>1\)
\(-a^2*b>-2a^2\)
We get \(0 > 1-2a^2\)
\(a^2 > 1/2\)
\(a^2 - (1/\sqrt{2})^2 > 0\)
\(a > 1/\sqrt{2}\) or a \(< -1/\sqrt{2}\)

Now, if you want to use the extreme value to guess the range, here is the logic for that:

Given:\(y>4\)
Find the range of x if \(x-y>-2\)
\(x > y-2\)
If y is a little more than 4, x is more than 2.
If y is 5, x is more than 3.
and so on...
The range that includes all these ranges is 'x is a little more than 2.'

Given: \(y>4\)
Find the range of x if \(x-2y<-6\)
\(x < 2y -6\)
If y is a little more than 4, x is less than something more than 2.
If y is 5, x is less than 4.
If y is 6, x is less than 6.
and so on...
The range that includes all these ranges is 'x is less than infinity.' which doesn't add much information. Algebraically, if you try to solve these two inequalities, you see you cannot get rid of y and hence don't get the range for x.

The diagram explains the above given two discussions.

Attachment:
Ques4.jpg
Ques4.jpg [ 7.55 KiB | Viewed 4531 times ]

Given: \(a^2*b>1\)
\(b< 2\)
If b is a little less than 2, a^2 is greater than 1/2 which means \(a > 1/\sqrt{2}\) or a \(< -1/\sqrt{2}\)
If b is 1, a^2 is greater than 1 which means \(a > 1\)or \(a < -1\)
b cannot be 0 or negative.
The range that includes all these ranges will be \(a > 1/\sqrt{2}\) or a \(< -1/\sqrt{2}\)
Attachment:
Ques5.jpg
Ques5.jpg [ 4.73 KiB | Viewed 4537 times ]
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Inequalities using extreme values. 21 Nov 2011, 08:05
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I wish I had the ability to give more than one kudos for a post! Thanks so much for your input! You should write "how to solve guides" for all problems and sell them!!!

you are so talented! when you describe something you make it seem SOOOO easy!

Really appreciate it!
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Inequalities using extreme values. 12 Oct 2015, 22:28
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