\(2x^2+3y^2-4x-12y+18=(2x^2-4x+2)+(3y^2-12y+12)+4=2(x-1)^2+3(y-2)^2+4\). 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.

Answer: C.
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Wow, you murdered this question... I found it to be strange, but you explained it perfectly. Kudos to you my friend!

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.

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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Nicely solved. I was wrong at my first attempt to solve this.

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

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.
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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

Hi Bunuel
please let me know the concept.
the operation is Ok but what is the logic behind such operation.
Thanks

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^(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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