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Find the minimum value of an expression 2 (x^2) + 3 (y^2) 
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13 Nov 2010, 14:11
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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
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Re: A strange question for me...
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13 Nov 2010, 14:20
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 \(2x^2+3y^24x12y+18=(2x^24x+2)+(3y^212y+12)+4=2(x1)^2+3(y2)^2+4\). Now, the least value of \(2(x1)^2\) and \(3(y2)^2\) is zero, so the least value of whole expression is 0+0+4=4. Answer: C.




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Re: A strange question for me...
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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!



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Re: Find the minimum value of an expression 2 (x^2) + 3 (y^2) 
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07 Feb 2014, 17:02
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(x1)^2 + 3 (y2)^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...
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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 \(2x^2+3y^24x12y+18=(2x^24x+2)+(3y^212y+12)+4=2(x1)^2+3(y2)^2+4\). Now, the least value of \(2(x1)^2\) and \(3(y2)^2\) is zero, so the least value of whole expression is 0+0+4=4. Answer: C. Wow!.. Amazing Solution... Bunuel, What should be the though process while breaking 18 to such forms ....



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Re: A strange question for me...
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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 \(2x^2+3y^24x12y+18=(2x^24x+2)+(3y^212y+12)+4=2(x1)^2+3(y2)^2+4\). Now, the least value of \(2(x1)^2\) and \(3(y2)^2\) is zero, so the least value of whole expression is 0+0+4=4. Answer: C. 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...
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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 \(2x^2+3y^24x12y+18=(2x^24x+2)+(3y^212y+12)+4=2(x1)^2+3(y2)^2+4\). Now, the least value of \(2(x1)^2\) and \(3(y2)^2\) is zero, so the least value of whole expression is 0+0+4=4. Answer: C. Nicely solved. I was wrong at my first attempt to solve this.



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Re: Find the minimum value of an expression 2 (x^2) + 3 (y^2) 
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16 Jul 2015, 09:59
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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Find the minimum value of an expression 2 (x^2) + 3 (y^2) 
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Updated on: 23 Apr 2016, 10:49
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 (xa)^2 or (yb)^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: \(2 (x^2) + 3 (y^2)  4x  12y + 18\) > \(2 (x^2) + 3 (y^2)  4x  12y + 2 + 16\) > \(2 (x^2) 4x + 2 + 3 (y^2)  12y + 16\) > \([2(x^2)  4x + 2]+ [3(y^2)12y + 12] +4\) > \(2(x1)^2 + 3 (y2)^2 + 4\) 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.



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Re: Find the minimum value of an expression 2 (x^2) + 3 (y^2) 
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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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Re: Find the minimum value of an expression 2 (x^2) + 3 (y^2) 
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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) 
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27 Feb 2017, 09:48
Answer: C
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^(n1) 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



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Re: Find the minimum value of an expression 2 (x^2) + 3 (y^2) 
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22 Mar 2018, 06:49
Spartan85 wrote: Answer: C
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^(n1) 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) 
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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) 
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