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# Find the minimum value of an expression 2 (x^2) + 3 (y^2) -

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Manager
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Joined: 25 Oct 2010
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Schools: Pitt, Oregon, LBS...
Find the minimum value of an expression 2 (x^2) + 3 (y^2) -  [#permalink]

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13 Nov 2010, 14:11
5
31
00:00

Difficulty:

65% (hard)

Question Stats:

63% (02:32) correct 37% (02:25) wrong based on 473 sessions

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Find the minimum value of an expression 2 (x^2) + 3 (y^2) - 4x - 12y + 18

A. 18
B. 10
C. 4
D. 0
E. -10
Math Expert
Joined: 02 Sep 2009
Posts: 59561
Re: A strange question for me...  [#permalink]

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13 Nov 2010, 14:20
17
5
MisterEko wrote:
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

$$2x^2+3y^2-4x-12y+18=(2x^2-4x+2)+(3y^2-12y+12)+4=2(x-1)^2+3(y-2)^2+4$$. Now, the least value of $$2(x-1)^2$$ and $$3(y-2)^2$$ is zero, so the least value of whole expression is 0+0+4=4.

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Manager
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Re: A strange question for me...  [#permalink]

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13 Nov 2010, 15:44
Wow, you murdered this question... I found it to be strange, but you explained it perfectly. Kudos to you my friend!
Intern
Joined: 01 Aug 2006
Posts: 31
Re: Find the minimum value of an expression 2 (x^2) + 3 (y^2) -  [#permalink]

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07 Feb 2014, 17:02
1
1
Completing the squares:
2 (x^2) + 3 (y^2) - 4x - 12y + 18 = 2(x^2 - 2x) + 3(y^2 - 4y) + 18
= 2(x^2 - 2x + 1) + 3 (y^2 - 4y + 4) + 18 +( -2 - 12)
= 2(x-1)^2 + 3 (y-2)^2 + 4

Minimum value is 4 because the least value the first two terms can take is ZERO.
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Re: A strange question for me...  [#permalink]

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07 Feb 2014, 21:20
Bunuel wrote:
MisterEko wrote:
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

$$2x^2+3y^2-4x-12y+18=(2x^2-4x+2)+(3y^2-12y+12)+4=2(x-1)^2+3(y-2)^2+4$$. Now, the least value of $$2(x-1)^2$$ and $$3(y-2)^2$$ is zero, so the least value of whole expression is 0+0+4=4.

Wow!.. Amazing Solution...

Bunuel, What should be the though process while breaking 18 to such forms ....
Math Expert
Joined: 02 Sep 2009
Posts: 59561
Re: A strange question for me...  [#permalink]

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10 Feb 2014, 01:14
Mountain14 wrote:
Bunuel wrote:
MisterEko wrote:
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

$$2x^2+3y^2-4x-12y+18=(2x^2-4x+2)+(3y^2-12y+12)+4=2(x-1)^2+3(y-2)^2+4$$. Now, the least value of $$2(x-1)^2$$ and $$3(y-2)^2$$ is zero, so the least value of whole expression is 0+0+4=4.

Wow!.. Amazing Solution...

Bunuel, What should be the though process while breaking 18 to such forms ....

The goal was to complete the squares for 2(x - ?)^2 and 3(y - ?)^2.
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Re: A strange question for me...  [#permalink]

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05 Mar 2014, 03:01
Bunuel wrote:
MisterEko wrote:
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

$$2x^2+3y^2-4x-12y+18=(2x^2-4x+2)+(3y^2-12y+12)+4=2(x-1)^2+3(y-2)^2+4$$. Now, the least value of $$2(x-1)^2$$ and $$3(y-2)^2$$ is zero, so the least value of whole expression is 0+0+4=4.

Nicely solved. I was wrong at my first attempt to solve this.
Intern
Joined: 16 Jul 2015
Posts: 1
Re: Find the minimum value of an expression 2 (x^2) + 3 (y^2) -  [#permalink]

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16 Jul 2015, 09:59
2
Alternative solution:

Within the expression 2(x^2) + 3(y^2) - 4x - 12y + 18 we have

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

By using the derivate for the first expression and putting it to 0 for minimum point we get

4x - 4 = 0
x = 1

Original expression 2(x^2) - 4x = -2

Same procedure for the second part

6y - 12 = 0
y = 2

Original expression 3(y^2) - 12y = -12

So now we have two parts of the expression which we have minimized and 18, which we can't do anything about. Putting it all together we get

(-2) + (-12) + 18 = 4
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Posts: 2561
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
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Find the minimum value of an expression 2 (x^2) + 3 (y^2) -  [#permalink]

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Updated on: 23 Apr 2016, 10:49
1
3
MisterEko wrote:
Find the minimum value of an expression 2 (x^2) + 3 (y^2) - 4x - 12y + 18

A. 18
B. 10
C. 4
D. 0
E. -10

For "minimum" questions, you will more often than not be able to create perfect squares by rearranging the terms. The benefit will be that by creating squares such as (x-a)^2 or (y-b)^2 you can get the minimum as 0 by just putting x =a or y =b. We do this as we are being asked about the minimum and not maximum.

Even for this question, with a bit of rearrangment we can come up with perfect squares are shown below:

$$2 (x^2) + 3 (y^2) - 4x - 12y + 18$$
---> $$2 (x^2) + 3 (y^2) - 4x - 12y + 2 + 16$$
----> $$2 (x^2) -4x + 2 + 3 (y^2) - 12y + 16$$
----> $$[2(x^2) - 4x + 2]+ [3(y^2)-12y + 12] +4$$
---> $$2(x-1)^2 + 3 (y-2)^2 + 4$$

Based on above, we can get minimum value by putting x=1 and y =2 to get 0 + 0 + 4 =4 . Thus C is the correct answer.

We put x=a or y =b as any perfect square will ALWAYS be $$\geq 0$$ . Thus the minimum value for any perfect square is 0.

Originally posted by ENGRTOMBA2018 on 16 Jul 2015, 10:18.
Last edited by ENGRTOMBA2018 on 23 Apr 2016, 10:49, edited 2 times in total.
Formatting.
Intern
Joined: 14 Mar 2016
Posts: 2
Re: Find the minimum value of an expression 2 (x^2) + 3 (y^2) -  [#permalink]

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29 May 2016, 13:49
I tried to solve in separate way .

ax2 +bx+c is has minimum value at -b/2a when a>0
2 (x^2) + 3 (y^2) - 4x - 12y + 18 can be written as 2(x^2)-4x+c & 3(y^2)-12y+c1.

1st equation is minimum at 4/2*2 = 1 , i.e min value of 1st equation is -2+c.

2nd equation is mimum at 12/3*2 = 2, i.e min value of 2nd equation is -12 +c1

Add above values -> -14+c+c1 is the minimum value
c+c1 is 18.

So the answer is 4. Option C
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Posts: 88
Re: Find the minimum value of an expression 2 (x^2) + 3 (y^2) -  [#permalink]

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03 Jun 2016, 15:34
Hi Bunuel
please let me know the concept.
the operation is Ok but what is the logic behind such operation.
Thanks
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Re: Find the minimum value of an expression 2 (x^2) + 3 (y^2) -  [#permalink]

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27 Feb 2017, 09:48
1
2

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

For extreme values of expression, d/dx(expression) =0

Differentiating with respect to x,
d/dx(expression) = 4x - 4= 0
Therefore, x= 1

Differentiating with respect to y,
d/dy(expression) = 6y - 12= 0
Therefore, y= 2

Substituting values x=1, y=2 in expression,
we get value = 4

It can be ascertained that this extreme value is minimum as d^2/dx^2(expression) = 4 (positive) and d^2/dy^2(expression) = 6 (positive)

Note: Differentiation is not tested on the GMAT, but following little information helps

1. d/dx(constant) = 0
2. d/dx(x^n) = n * x^(n-1)
3. d/dx(c*x) = c ; where c = constant
4. Expression has extreme values at d/dx(expression)=0
; d2/dx^2(expression) = positive signifies minimum value
; d2/dx^2(expression) = negative signifies maximum value
Senior Manager
Joined: 17 Mar 2014
Posts: 426
Re: Find the minimum value of an expression 2 (x^2) + 3 (y^2) -  [#permalink]

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22 Mar 2018, 06:49
Spartan85 wrote:

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

For extreme values of expression, d/dx(expression) =0

Differentiating with respect to x,
d/dx(expression) = 4x - 4= 0
Therefore, x= 1

Differentiating with respect to y,
d/dy(expression) = 6y - 12= 0
Therefore, y= 2

Substituting values x=1, y=2 in expression,
we get value = 4

It can be ascertained that this extreme value is minimum as d^2/dx^2(expression) = 4 (positive) and d^2/dy^2(expression) = 6 (positive)

Note: Differentiation is not tested on the GMAT, but following little information helps

1. d/dx(constant) = 0
2. d/dx(x^n) = n * x^(n-1)
3. d/dx(c*x) = c ; where c = constant
4. Expression has extreme values at d/dx(expression)=0
; d2/dx^2(expression) = positive signifies minimum value
; d2/dx^2(expression) = negative signifies maximum value

wow..thank you very much. Above method made my life easy
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Re: Find the minimum value of an expression 2 (x^2) + 3 (y^2) -  [#permalink]

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10 Jul 2019, 12:48
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Re: Find the minimum value of an expression 2 (x^2) + 3 (y^2) -   [#permalink] 10 Jul 2019, 12:48
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