C the answer.

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

2*(x^2 -2x +1) + 3*(y^2 -4y + 4) + (18 - 2 - 12)

2*(x-1)^2 + 3*(y-2)^2 + 4

The minimum value of this expression is 4 at it is when, x=1, y=2

next time, for the questions like this. it's factored it to square, and don't care it and look for the remain (ex: it's 4). it's the answer.
you will save your time. i've done it without paper just 15s. teeheee

answering to the original q. in this thread
d/dx=4x-4=0, x=1 and d/dy=6y-12=0, y=2
d^2/dx^2=4 (implies min x) and d^2/dy^2=6 (implies min y)
min f(x,y) = 2*(1^2) +3*(2^2)-4*1-12*2+18=4 and choice C

Using multivariate calculus turns the solution of this q. into joke

winnerhere 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

How about -b/2a being the minimum value (maximum in certain contexts) of a quadratic expression (or equation).
For the one in x: 4/2(2)=1
For the one in y: 12/2(3)=3
Summing it 4 that forms the minimum value of the total expression

No?

pemdas wrote:Using multivariate calculus turns the solution of this q. into joke

... which is exactly why THIS QUESTION WOULD NOT BE ON THE GMAT. Questions that become routine with "advanced" formulas are not asked, period.

I like that people are posting assorted math questions here, but it's irresponsible to present these as GMAT questions. Students have enough material to worry about already!

neerajkumar1_1 wrote:I suppose rahul gave an excellent answer by clubbing the x and y to form perfect squares and the making them 0..

If you want to be doubly sure that u got the minimum value,
there is a concept of differentiation...

All it says is that an expressions value will be min or max when u differentiate it for the variable in question and equate it to 0..
so when we differentiate the expression with respect to x (treating any other variable as constant), we get
4x=4
x=1
and when we differentiate the expression with respect to y (treating any other variable as constant), we get
6y=12
y=2
put these values back in the expression and u will get value of expression equal to 4

Pick C...

ps: the logic overall remains the same...
we basically try to find values for x and y, such that the value of exp is minimum...

The above differentiation method CAN give the maxima or minima of an expression, so to check if differentiation method IS giving us minima or maxima - we need to do the double differentiation as well.
If double differentiation brings a positive value - then the single differntiation will give a Minima conversely if the double differentiation brings a negative value - then the single differentiation will provide a Maxima.
In this case the double differentiation of the expression - f'(x)=4x-4 => is f''(x)=4 - which is always positive. so the single differentiation of the expression will provide a Minima (minimum value).
From GMAT perspective, we are not supposed to know all this; and the best method would be to go for perfect square method to find out the minimum value of expression as shown by one of the members already.

Assume x = ~0 and y = ~0

May I point out a small error in this argument:

2x (x-2) = 0* -2 = 0 (not -4)
3y (y-4) = 0* -4 = 0 (not -12 )

So the answer would be higher => 0 + 0 + 18 = 18

Remember that multiplying by zero = zero.

goyalsau wrote:

thebigkats wrote:out of curiosity - problem doesn;t state that x and y are integers.
so x and y could have been extremely close to 0 +ve numbers which could take us to a different answer altogether - 2
Assume x = ~0 and y = ~0

2x (x-2) = ~2* -2 = ~-4
3y (y-4) = ~3* -4 =~-12

and the answer is => ~-4 + ~-12 + 18 = ~2

Should we assume that variables are integers in problems like this?

Nice Thought, But i think Gmat will not left the room for arguments,
If this would have been a Gmat question they would have particularly mentioned about x and y,
AS per my experience,

I read somewhere that gmat spends more than 700 $ on a single question. ........

= 2(x² - 2x + 1) + 3(y² - 4y + 4) + 4
= 2(x - 1)² + 3(y - 2)² + 4

I dont get this part. How do we get the values +1 and +4 and what happens to the +18. I seem to be missing it. Can someone help me out?

Separating the negative and positive parts, we get:

Find the minimum value of A + B + 18 when

A = 2x^2 + 3y^2
B = - 4x - 12y

As the variables in A are squared, then A will always return a positive answer.
As the variables in B both have a negative coefficient, then x>0 and y>0 to minimise B (as -+ = -).
INSUFFICIENT INFO.

Now treat x and y separately:
2x^2 - 4x and 3y^2 - 12y both need to be minimum.

Therefore:
x^2 - 2x and y^2 - 4y both need to be minimum.

Looking at x first:
X = x^2 - 2x
dX/dx = 2x - 2 = 0 when x = 1
d2X/dx = 2 > 0 therefore MINIMUM
So X min = 1 - 2 = -1

Looking at y:
Y = y^2 - 4y
dY/dx = 2y - 4 = 0 when y = 2
d2Y/dX2 = 2 > 0 therefore MINIMUM
So, Y min = 4 - 8 = -4

Substitituting (x,y) = (1,2) gives

2 (1) + 3 (2^2) - 4 - 12(2) + 18 = 2 + 12 - 4 - 24 + 18 = 4

So minimum value is answer (3).

ans is a ie 18

Great to see the enthusiasm here, but this thread needs to die - this is a question for introductory calculus, NOT the GMAT! There are thousands of other questions that are more worth your time to review.

let z=2*(x^2) + 3*(y^2) - 4x -12y + 18. then for minimum
dz/dx=0
d(2*(x^2) + 3*(y^2) - 4x -12y + 18)/dx=0
dz/dx=4x-4=0
x=1
dz/dy=0
d(2*(x^2) + 3*(y^2) - 4x -12y + 18)/dy=0
dz/dy=6y-12=0
y=2

by putting x=1 and y=2 in z=2*(x^2) + 3*(y^2) - 4x -12y + 18
we get z=4

so my answer is z=4

Brent@GMATPrepNow wrote:Okay - no more calculus solutions. This question is out of scope for the GMAT.
Let it go.

Cheers,
Brent

Sometimes I wonder if someone, somewhere, is training computers to write answers to problems and is unleashing those computers on these message boards. (That's at least one explanation for how often we get someone reviving an long-dead thread with a less helpful (and often incorrect!) explanation than the ones that have already been posted.)