Minimum value of an expression

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Minimum value of an expression

by winnerhere » Mon Jul 19, 2010 7:03 pm
Find the minimum value of an expression

2 (x^2) + 3 (y^2) - 4x - 12y + 18

1) 18

2)10

3) 4

4) 0

5) -10

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by [email protected] » Mon Jul 19, 2010 7:21 pm
Solution:
The above expression is same as 2*{(x-1)^2} + 3*{(y - 2)^2} + 4.
Now the minimum value of the expression will be when x is 1 and y is 2.
Then the square terms containing x and y are zero and we are left with the constant term 4.
Or 4 is the minimum value.
The correct answer is hence 3).
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by winnerhere » Mon Jul 26, 2010 6:59 pm
Thanks Rahul :)

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by guateflava » Mon Jul 26, 2010 7:27 pm
I'm sorry, I don't get it. How do you get from the stated formula to your result? I must be forgetting some rule.

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by guateflava » Mon Jul 26, 2010 7:37 pm
I'm sorry, I don't get it. How do you get from the stated formula to your result? I must be forgetting some rule.

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by mbaonmind » Mon Jul 26, 2010 10:19 pm
There is one way called differentiation,

the expression is 2x^2 +3y^2-4x-12y+18

If I differentiate with respect x and equate to 0 i get
4x-4 = 0
i.e. x = 1;

If I differentiate with respect y and equate to 0 i get
6y-12 = 0
i.e. y = 2

when substituted x = 1, y = 2 we get expression value as 4 Thus Answer is (3)

I do not know any other way to get values for x and y except Differentiation or hit/try by substituting values. and the later is very time consuming.

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by neerajkumar1_1 » Sat Nov 13, 2010 6:20 pm
I suppose rahul gave an excellent answer by clubbing the x and y to form perfect squares and the making them 0..

If you want to be doubly sure that u got the minimum value,
there is a concept of differentiation...

All it says is that an expressions value will be min or max when u differentiate it for the variable in question and equate it to 0..
so when we differentiate the expression with respect to x (treating any other variable as constant), we get
4x=4
x=1
and when we differentiate the expression with respect to y (treating any other variable as constant), we get
6y=12
y=2
put these values back in the expression and u will get value of expression equal to 4

Pick C...

ps: the logic overall remains the same...
we basically try to find values for x and y, such that the value of exp is minimum...

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by goyalsau » Sun Nov 14, 2010 9:53 pm
neerajkumar1_1 wrote:I suppose rahul gave an excellent answer by clubbing the x and y to form perfect squares and the making them 0..

If you want to be doubly sure that u got the minimum value,
there is a concept of differentiation...

All it says is that an expressions value will be min or max when u differentiate it for the variable in question and equate it to 0..
so when we differentiate the expression with respect to x (treating any other variable as constant), we get
4x=4
x=1
and when we differentiate the expression with respect to y (treating any other variable as constant), we get
6y=12
y=2
put these values back in the expression and u will get value of expression equal to 4

Pick C...

ps: the logic overall remains the same...
we basically try to find values for x and y, such that the value of exp is minimum...
Nice work ,
If it asked for the largest value of the expression.
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by thp510 » Sun Nov 14, 2010 11:49 pm
neerajkumar1_1 wrote: If you want to be doubly sure that u got the minimum value,
there is a concept of differentiation...

All it says is that an expressions value will be min or max when u differentiate it for the variable in question and equate it to 0..
so when we differentiate the expression with respect to x (treating any other variable as constant), we get
4x=4
x=1
and when we differentiate the expression with respect to y (treating any other variable as constant), we get
6y=12
y=2
put these values back in the expression and u will get value of expression equal to 4
Not getting it. What does the equation look like before you get to 4x=4 or 6y=12 using the differentiation rule. I'm still stuck at the following:

2(x^2-2x)+3(y^2-4y)+18

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by [email protected] » Mon Nov 15, 2010 1:35 am
thp510 wrote:...
Not getting it. What does the equation look like before you get to 4x=4 or 6y=12 using the differentiation rule. I'm still stuck at the following:

2(x^2-2x)+3(y^2-4y)+18
To use the differentiation rule, you need to know differential calculus which is not a part of GMAT. You can proceed as I mentioned earlier. Let's do it again. To minimize the expression, we have to minimize each term of it. Simple algebraical method to do it is to rearrange the terms in such a way that we get square terms. This is because minimum value of a square term is zero.

# 2x² + 3y² - 4x - 12y + 18
= 2(x² - 2x) + 3(y² - 4y) + 18 .................... What you've done
= 2(x² - 2x + 1) + 3(y² - 4y + 4) + 4
= 2(x - 1)² + 3(y - 2)² + 4

For minimum value of the expression, 2(x - 1)² and 3(y - 2)² must be minimum, i.e. equal to zero. Thus minimum value of the expression is 4.
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by thp510 » Mon Nov 15, 2010 8:01 am
[email protected] wrote:
thp510 wrote:...
Not getting it. What does the equation look like before you get to 4x=4 or 6y=12 using the differentiation rule. I'm still stuck at the following:

2(x^2-2x)+3(y^2-4y)+18
To use the differentiation rule, you need to know differential calculus which is not a part of GMAT. You can proceed as I mentioned earlier. Let's do it again. To minimize the expression, we have to minimize each term of it. Simple algebraical method to do it is to rearrange the terms in such a way that we get square terms. This is because minimum value of a square term is zero.

# 2x² + 3y² - 4x - 12y + 18
= 2(x² - 2x) + 3(y² - 4y) + 18 .................... What you've done
= 2(x² - 2x + 1) + 3(y² - 4y + 4) + 4
= 2(x - 1)² + 3(y - 2)² + 4

For minimum value of the expression, 2(x - 1)² and 3(y - 2)² must be minimum, i.e. equal to zero. Thus minimum value of the expression is 4.
Interesting. Thanks for the new math info! :)

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by Abhishek009 » Thu Nov 18, 2010 2:20 am
[email protected] wrote:
thp510 wrote:...
Not getting it. What does the equation look like before you get to 4x=4 or 6y=12 using the differentiation rule. I'm still stuck at the following:

2(x^2-2x)+3(y^2-4y)+18
To use the differentiation rule, you need to know differential calculus which is not a part of GMAT. You can proceed as I mentioned earlier. Let's do it again. To minimize the expression, we have to minimize each term of it. Simple algebraical method to do it is to rearrange the terms in such a way that we get square terms. This is because minimum value of a square term is zero.

# 2x² + 3y² - 4x - 12y + 18
= 2(x² - 2x) + 3(y² - 4y) + 18 .................... What you've done
= 2(x² - 2x + 1) + 3(y² - 4y + 4) + 4
= 2(x - 1)² + 3(y - 2)² + 4

For minimum value of the expression, 2(x - 1)² and 3(y - 2)² must be minimum, i.e. equal to zero. Thus minimum value of the expression is 4.
Well Rahul you are absolutely correct but I feel a little of knowledge on Differrentiation is not that hard to learn and master , infact it will give the test takers an added advantage to solve such problems.

Here is the link for those who are interested in differentiation :

https://www.statistica.com.au/differenti ... d_min.html
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by thebigkats » Thu Nov 18, 2010 12:31 pm
out of curiosity - problem doesn;t state that x and y are integers.
so x and y could have been extremely close to 0 +ve numbers which could take us to a different answer altogether - 2
Assume x = ~0 and y = ~0

2x (x-2) = ~2* -2 = ~-4
3y (y-4) = ~3* -4 =~-12

and the answer is => ~-4 + ~-12 + 18 = ~2

Should we assume that variables are integers in problems like this?

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by goyalsau » Thu Nov 18, 2010 7:48 pm
thebigkats wrote:out of curiosity - problem doesn;t state that x and y are integers.
so x and y could have been extremely close to 0 +ve numbers which could take us to a different answer altogether - 2
Assume x = ~0 and y = ~0

2x (x-2) = ~2* -2 = ~-4
3y (y-4) = ~3* -4 =~-12

and the answer is => ~-4 + ~-12 + 18 = ~2

Should we assume that variables are integers in problems like this?

Nice Thought, But i think Gmat will not left the room for arguments,
If this would have been a Gmat question they would have particularly mentioned about x and y,
AS per my experience,

I read somewhere that gmat spends more than 700 $ on a single question. ........
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by [email protected] » Fri Nov 19, 2010 1:45 am
Hi there,

I will NOT go into mathematical details here, because I guess it would not be GMAT-focused, but I guess it´s my "obligation" to make some very-short considerations:

01. Rahul´s solution is not only good for GMAT´s purpose, it´s mathematically impecable not only as arguments are concerned, but also as the RIGHT ANSWER is obtained.

I guess many posts are trying to avoid it, because it seems "artificial", but it is simply a "completing the squares" traditional method.

Let me show you how you could find his "simplified" (equivalent) expression without "trial-and-error"...

2(x^2 - 2x) + 3(y^2 - 4y) + 18 = 2(x^2 - 2x +1) -2 + 3(y^2 - 4y +4) - 12 + 18 = 2(x-1)^2 + 3 (y-2)^2 -2-12+18 and the next passage is the one Rahul´s found! Simple as that.


02. The posts that are trying to use Calculus are trying to find minimum/maximum values of 2-real variable functions without care, and they are using wrong reasoning arguments. It´s is NOT true, in general, that you can find extreme values leaving one parameter fixed, then the other, etc... that´s why conditions of minimality/maximality usually considers some matrix/determinants to find points of inflexions, etc etc etc.

03. The post that deals with approximations is also non-rigorous and, may I put directly (no pun intended), simple wrong.
Assume x = ~0 and y = ~0

2x (x-2) = ~2* -2 = ~-4
3y (y-4) = ~3* -4 =~-12

and the answer is => ~-4 + ~-12 + 18 = ~2
Are you kidding me? If x is near zero, you say 2x (x-2) is approx. -4 ??

Please note that 2x(x-2) = 2x^2 - 4x and when x goes to zero, both 2x^2 and 4x go too, therefore 2x(x-2) does not approach -4 when x approaches zero!! Think about the parabola y = 2x^2 - 4x, the x-vertex is zero and it passes through the origin (0,0) and its concavity is positive, therefore when x goes near zero, the points (x, 2x(x-2)) goes near the point (0,0), got it?

Well, in short: you don´t need to have x and y integers to have 4 as the minimum of the expression given and please do not drive (I mean "use Mathematics") after drinking (I´m joking, no pun intended)!!

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Fabio.
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