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Minimum Value

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Source: — Problem Solving |
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by vitaly » Fri May 08, 2009 7:16 pm
We have function f(x, y) = 3x2 – 6xy + 4y2 – 4y + 11

So let's take partial derivatives:
f'x(x,y) = 6x-6y
f'y(x,y) = -6x+8y-4

=> min (or max) is in point (x, y) where x and y match these conditions:
(1) 6x-6y = 0
(2) -6x+8y-4 = 0

from (1) we have x=y, and then from (2) that x=y=2

f(2,2)=3*4-6*4+4*4-4*2+11=4-8+11=7

It's min, since f(0,0)=11 > 7.

So answer is e)
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by Vemuri » Fri May 08, 2009 11:35 pm
vitaly wrote:We have function f(x, y) = 3x2 – 6xy + 4y2 – 4y + 11

So let's take partial derivatives:
f'x(x,y) = 6x-6y
f'y(x,y) = -6x+8y-4
Can you please explain how you derived the partial derivatives for the expression 3x^2 – 6xy + 4y^2 – 4y + 11
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Re: Minimum Value

by sureshbala » Fri May 08, 2009 11:42 pm
dtweah wrote:What is the minimum value, for all real numbers x and y, of the expression 3x2 – 6xy + 4y2 – 4y + 11?

(a) -1
(b) 2
(c) 9/2
(d) 8/3
(e) 7
Folks, there's no need of derivatives to solve this and this topic is not meant for GMAT aspirants....

Let us crack this with elementary math ....

3x^2 - 6xy + 4y^2 -4y + 11

= 3x^2 - 6xy + 3y^2 + y^2 - 4y + 11

= 3(x^2 - 2xy + y^2) + y^2 - 4y + 11

= 3(x - y)^2 + (y - 2)^2 + 7.

Now (x - y)^2 > = 0 and (y - 2)^2 >= 0, the above expression will have minimum value only when (x - y) =0 and (y - 2 ) = 0.

Hence the minimum value is 7.

(When u face such questions in the exam all you have to do is to express it as a combination of perfect squares and the constant left will be the minimum value of the given expression)
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Re: Minimum Value

by Vemuri » Sat May 09, 2009 1:48 am
sureshbala wrote: Folks, there's no need of derivatives to solve this and this topic is not meant for GMAT aspirants....

Let us crack this with elementary math ....

3x^2 - 6xy + 4y^2 -4y + 11

= 3x^2 - 6xy + 3y^2 + y^2 - 4y + 11

= 3(x^2 - 2xy + y^2) + y^2 - 4y + 11

= 3(x - y)^2 + (y - 2)^2 + 7.

Now (x - y)^2 > = 0 and (y - 2)^2 >= 0, the above expression will have minimum value only when (x - y) =0 and (y - 2 ) = 0.

Hence the minimum value is 7.

(When u face such questions in the exam all you have to do is to express it as a combination of perfect squares and the constant left will be the minimum value of the given expression)
perfecto !!! Thank you for such a lovely explanation.
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by vitaly » Sat May 09, 2009 5:52 am
>Can you please explain how you derived the partial derivatives for the expression 3x^2 – 6xy + 4y^2 – 4y + 11

It's easy, you should have a variable (for example, you need derivative by x). Then you consider y as a constant.

So it will be
(3x^2)' = 6x
(-6xy)'= -6y
rest is 0.

Same thing you do condidering y as variable, x as constant.
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