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M28-10

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M28-10  [#permalink]

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New post 16 Sep 2014, 00:28
1
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A
B
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D
E

Difficulty:

  95% (hard)

Question Stats:

37% (01:13) correct 63% (01:46) wrong based on 178 sessions

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Re M28-10  [#permalink]

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New post 16 Sep 2014, 00:28
Official Solution:

What is the maximum value of \(-3x^2 + 12x -2y^2 - 12y - 39\) ?

A. \(-39\)
B. \(-9\)
C. \(0\)
D. \(9\)
E. \(39\)


\(-3x^2 + 12x -2y^2 - 12y - 39=-3x^2 + 12x-12-2y^2 - 12y-18-9=\)

\(-3(x^2-4x+4)-2(y^2+6y+9)-9=\)

\(=-3(x-2)^2-2(y+3)^2-9\).

So, we need to maximize the value of \(-3(x-2)^2-2(y+3)^2-9\).

Since, the maximum value of \(-3(x-2)^2\) and \(-2(y+3)^2\) is zero, then the maximum value of the whole expression is \(0+0-9=-9\).


Answer: B
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Re: M28-10  [#permalink]

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New post 15 Oct 2015, 11:04
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How did you know to split the -39 into 3 parts: -12, -18, and -9?
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Re: M28-10  [#permalink]

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New post 05 Mar 2016, 17:14
iwantstanford wrote:
How did you know to split the -39 into 3 parts: -12, -18, and -9?


Hi Experts - can you please clarify how to recognize that we need to split -39 to solve?

I'm not sure how to recognize that this is the approach. Also, how do you know what to split -39 into?
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M28-10  [#permalink]

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New post 05 Mar 2016, 17:37
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happyface101 wrote:
iwantstanford wrote:
How did you know to split the -39 into 3 parts: -12, -18, and -9?


Hi Experts - can you please clarify how to recognize that we need to split -39 to solve?

I'm not sure how to recognize that this is the approach. Also, how do you know what to split -39 into?


The trick with these questions that ask you about max/min of a second degree polynomial is that you will invariably get a value of the form (a\(\pm\)b) wherein, the minimum value of any square = 0. Thus try to create perfect squares out of the given polynomials.


Realize that \((a\pm b)^2 = a^2 \pm 2*a*b + b^2\).

Given expression: \(−3x^2+12x−2y^2−12y−39\) ---> try to create perfect squares :

\((-3x^2+12x) + (-2y^2-12y)-39 ---> -3 (x^2-4x) -2 (y^2+6y) -39 ----> 3 (x^2-4x+4-4) -2 (y^2+6y+9-9) -39\)


---> \(3 (x^2-4x+4-4) -2 (y^2+6y+9-9) -39\) ---> \(-3[(x-2)^2-4] -2 [(y+3)^2-9]-39\) ----> \(-3(x-2)^2+12 -2(y+3)^2+18-39\)

----> \(-3(x-2)^2+12 -2(y+3)^2+18-39\) --->\(-3(x-2)^2-2(y+3)^2-9\)

As mentioned above, the minimum value of a perfect square = 0 --> to maximize the given expression,

you need to put \((x-2)^2 =0\) and\((y+3)^2=0\) ---> giving you 0+0-9=-9 as the maximum value.

Any other values of\((x-3)^2\) and \((y+3)^2\)will give you a smaller negative value hence 'minimizing' instead of maximizing the expression.

Hope this helps.
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Re: M28-10  [#permalink]

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New post 14 Mar 2016, 04:03
Bunuel wrote:
Official Solution:

What is the maximum value of \(-3x^2 + 12x -2y^2 - 12y - 39\) ?

A. \(-39\)
B. \(-9\)
C. \(0\)
D. \(9\)
E. \(39\)


\(-3x^2 + 12x -2y^2 - 12y - 39=-3x^2 + 12x-12-2y^2 - 12y-18-9=\)

\(-3(x^2-4x+4)-2(y^2+6y+9)-9=\)

\(=-3(x-2)^2-2(y+3)^2-9\).

So, we need to maximize the value of \(-3(x-2)^2-2(y+3)^2-9\).

Since, the maximum value of \(-3(x-2)^2\) and \(-2(y+3)^2\) is zero, then the maximum value of the whole expression is \(0+0-9=-9\).


Answer: B

'Hi
Is there a quick way to find the split?
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Re: M28-10  [#permalink]

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New post 14 Mar 2016, 04:06
sinhap07 wrote:
Bunuel wrote:
Official Solution:

What is the maximum value of \(-3x^2 + 12x -2y^2 - 12y - 39\) ?

A. \(-39\)
B. \(-9\)
C. \(0\)
D. \(9\)
E. \(39\)


\(-3x^2 + 12x -2y^2 - 12y - 39=-3x^2 + 12x-12-2y^2 - 12y-18-9=\)

\(-3(x^2-4x+4)-2(y^2+6y+9)-9=\)

\(=-3(x-2)^2-2(y+3)^2-9\).

So, we need to maximize the value of \(-3(x-2)^2-2(y+3)^2-9\).

Since, the maximum value of \(-3(x-2)^2\) and \(-2(y+3)^2\) is zero, then the maximum value of the whole expression is \(0+0-9=-9\).


Answer: B

'Hi
Is there a quick way to find the split?


No, there is no "quickest" way than the ones mentioned above. With practice, this should not take you more than 2 minutes.
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Re: M28-10  [#permalink]

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New post 15 Aug 2016, 13:02
I'm confused how you got to this

(−3x 2 +12x)+(−2y 2 −12y)−39−−−>−3(x 2 −4x)−2(y 2 +6y)−39−−−−>3(x 2 −4x+4−4)−2(y 2 +6y+9−9)−39
(−3x2+12x)+(−2y2−12y)−39−−−>−3(x2−4x)−2(y2+6y)−39−−−−>3(x2−4x+4−4)−2(y2+6y+9−9)−3

I see what you factoried out the 2 and the -2, althought I'm not sure why the 2X and -2y were't factored out.

I'm lose how we got from 3(x²-4x) to 3(x²-4x+4-4) (i understand that it is the same thing, but why break it down like that?)

Kind of lost.

Thanks
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Re: M28-10  [#permalink]

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New post 19 Feb 2017, 14:59
1
Bunuel wrote:
Official Solution:

What is the maximum value of \(-3x^2 + 12x -2y^2 - 12y - 39\) ?

A. \(-39\)
B. \(-9\)
C. \(0\)
D. \(9\)
E. \(39\)


\(-3x^2 + 12x -2y^2 - 12y - 39=-3x^2 + 12x-12-2y^2 - 12y-18-9=\)

\(-3(x^2-4x+4)-2(y^2+6y+9)-9=\)

\(=-3(x-2)^2-2(y+3)^2-9\).

So, we need to maximize the value of \(-3(x-2)^2-2(y+3)^2-9\).

Since, the maximum value of \(-3(x-2)^2\) and \(-2(y+3)^2\) is zero, then the maximum value of the whole expression is \(0+0-9=-9\).


Answer: B


Bunuel, I'm sure you know this already but for this question a very quick shortcut would be Calculus, although obviously this isn't officially tested on the GMAT. I wanted to ask if you have you ever come across a Max/Min question where Calculus is a bad idea? Just want to make sure I don't fall into a trap, get caught in a problem that eats up a ton of time, etc.
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Re: M28-10  [#permalink]

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New post 19 Feb 2017, 21:26
brooklyndude wrote:
Bunuel wrote:
Official Solution:

What is the maximum value of \(-3x^2 + 12x -2y^2 - 12y - 39\) ?

A. \(-39\)
B. \(-9\)
C. \(0\)
D. \(9\)
E. \(39\)


\(-3x^2 + 12x -2y^2 - 12y - 39=-3x^2 + 12x-12-2y^2 - 12y-18-9=\)

\(-3(x^2-4x+4)-2(y^2+6y+9)-9=\)

\(=-3(x-2)^2-2(y+3)^2-9\).

So, we need to maximize the value of \(-3(x-2)^2-2(y+3)^2-9\).

Since, the maximum value of \(-3(x-2)^2\) and \(-2(y+3)^2\) is zero, then the maximum value of the whole expression is \(0+0-9=-9\).


Answer: B


Bunuel, I'm sure you know this already but for this question a very quick shortcut would be Calculus, although obviously this isn't officially tested on the GMAT. I wanted to ask if you have you ever come across a Max/Min question where Calculus is a bad idea? Just want to make sure I don't fall into a trap, get caught in a problem that eats up a ton of time, etc.


You can solve GMAT questions without calculus. So, no need to study it for the purpose of the test. On the other hand if you know that part of marth it won't hurt.
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M28-10  [#permalink]

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New post 22 Oct 2018, 19:50
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If you are familiar with basic differentiation, it would be very quick.

Solution: Let the given equation be 1.

Step 1.Find differentiation of equ 1 wr.to "X" and equate to "0" (This is the standard rule for a differentiation in order to get maximum value)
Hence, -6x+12=0
therefore x=2

Step 2.Then differentiate the equ 1 wr.to "Y" and equate to "0"
-4y-12=0
therefore y=-3

Step 3.Submit both x=2 and y=-3 in the equ. 1 to get the maximum value.
Therefore the answer is -9.

Differentiation approach can save lot of time while determining the maximum value of an expression.
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Re: M28-10  [#permalink]

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New post 23 Oct 2018, 08:14
to create perfect square u need to add and subtract (-b/2a)^2 in the equation ax^2 + bx + c = 0
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Re: M28-10 &nbs [#permalink] 23 Oct 2018, 08:14
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