The minimum value is 4 and it is possible when x=1 and y=2
Hence C is the answer
Minimum value of an expression
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akash singhal
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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.
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.
Can you please explain how the 18 became 4 in that equation??
Rahul@gurome wrote: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.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
# 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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DavidG@VeritasPrep
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Rahul doesn't seem to have been active on the site for some time. If you look at the third line:christitk wrote:Can you please explain how the 18 became 4 in that equation??Rahul@gurome wrote: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.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
# 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.
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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Matt@VeritasPrep
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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 
- but AFAIK it's totally irrelevant to the GMAT, so you can safely ignore this question for the time being.
- but AFAIK it's totally irrelevant to the GMAT, so you can safely ignore this question for the time being.-
deepak4mba
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