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Minimum value of an expression

Expert replies

by akash singhal » Mon May 11, 2015 9:49 pm
The minimum value is 4 and it is possible when x=1 and y=2

Hence C is the answer
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by mbatious » Mon May 30, 2016 11:21 pm
I tried a slightly different method.

Lets say we substitute x=1 and y=1. We get 7. Minimum value cant be 18 and 10, eliminate A and B.
For getting min value negative x and y have to be negative, their squares will be much higher than their face value, eliminate -10.

0 cant be a solution because you cant solve the equation for 0.

We are left with 4 (C).

I solved this in less than a minute.
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by christitk » Tue Aug 15, 2017 10:26 am
Can you please explain how the 18 became 4 in that equation??
Rahul@gurome 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.
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by DavidG@VeritasPrep » Tue Aug 15, 2017 10:34 am
christitk wrote:Can you please explain how the 18 became 4 in that equation??
Rahul@gurome 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.
Rahul doesn't seem to have been active on the site for some time. If you look at the third line:

2(x² - 2x + 1) + 3(y² - 4y + 4) + 4 = 2x² - 4x + 2 + 3y² - 12y + 12 + 4

The terms in red sum to 18. He broke it into 2, 12, and 4 for purposes of creating square terms.
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by Matt@VeritasPrep » Fri Aug 18, 2017 2:08 pm
Completing the square is a very clever idea - I remember how excited I was the first time I saw it, and I won't apologize for that :D - but AFAIK it's totally irrelevant to the GMAT, so you can safely ignore this question for the time being.
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by deepak4mba » Sun Mar 04, 2018 4:07 am
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