If x and y are real numbers, then the minimum value of x^2+4xy+6y^2-4y

Question

If x and y are real numbers, then the minimum value of  x^2+4xy+6y^2-4y+4  is?

A. -4

B. 0

C. 2

D. 3

E. 4

dotsoftme · Answer

To find the minimum value of the expression $x^2 + 4xy + 6y^2 - 4y + 4$ for real numbers $x$ and $y$, we can complete the square for both variables.
First, let's group the terms involving $x$ to complete the square for $x$:
$$x^2 + 4xy + 6y^2 - 4y + 4$$
We can split $6y^2$ into $4y^2 + 2y^2$:
$$= (x^2 + 4xy + 4y^2) + 2y^2 - 4y + 4$$
The first group forms a perfect square:
$$= (x + 2y)^2 + 2y^2 - 4y + 4$$
Next, we complete the square for the remaining terms involving $y$:
$$2y^2 - 4y + 4 = 2(y^2 - 2y) + 4$$
$$= 2(y^2 - 2y + 1 - 1) + 4$$
$$= 2(y - 1)^2 - 2 + 4$$
$$= 2(y - 1)^2 + 2$$
Now substitute this back into the expression:
$$= (x + 2y)^2 + 2(y - 1)^2 + 2$$
Since $x$ and $y$ are real numbers, the squared terms $(x + 2y)^2$ and $(y - 1)^2$ are always greater than or equal to zero:
$$(x + 2y)^2 \ge 0$$
$$(y - 1)^2 \ge 0$$
Therefore, the entire expression is minimized when both squared terms equal zero:
 
 $y - 1 = 0 \implies y = 1$ 
 $x + 2y = 0 \implies x + 2(1) = 0 \implies x = -2$  
When $x = -2$ and $y = 1$, the minimum value of the expression is:
$$0 + 2(0) + 2 = 2$$
 Correct Option:C. 2

modiassumenda · Answer

We want the minimum value of:
x2 + 4xy + 6y2 − 4y + 4
Step 1: Complete the square for x.
x2 + 4xy + 6y2 − 4y + 4
= (x + 2y)2 + 2y2 − 4y + 4
because
(x + 2y)2 = x2 + 4xy + 4y2.
Step 2: Complete the square for y.
2y2 − 4y + 4
= 2(y2 − 2y) + 4
= 2  + 4
= 2(y − 1)2 + 2
So the expression becomes:
(x + 2y)2 + 2(y − 1)2 + 2
Step 3: Find the minimum.
Both (x + 2y)2 and 2(y − 1)2 are always non-negative.
Therefore, the smallest value occurs when:
x + 2y = 0
and
y = 1
This gives:
x = −2
and
y = 1
Substituting these values makes both square terms 0, so the minimum value is: 2
 Answer: 2  ✅

gmataccount123 · Answer

Substituting $x = -2$ and $y = 1$ yields the minimum value:$0+2(0)+2=2$ how you get it

egmat · Answer

Hi gmataccount123,

Your algebra is fine; the only thing tripping you up is reading the " 0  +  2(0)  +  2 " line, so let me decode it term by term.

The expression was rewritten as three pieces:

(x + 2y)2 + 2(y - 1)2 + 2

Now plug in x =  -2  and y =  1  into each piece separately:

-  First piece - (x + 2y)2:  x + 2y = -2 + 2(1) =  0 , and 0 squared =  0 . That's the leading  0 .
-  Second piece - 2(y - 1)2:  y - 1 = 1 - 1 =  0 , so 2 x 0 squared = 2 x 0 =  0 . That's the  2(0)  you saw - the " 2 " is just the coefficient sitting outside the square, and the square itself is  0 .
-  Third piece - the +2:  this is a plain constant. Nothing gets substituted into it, so it stays  2 .

Add them up:  0  +  0  +  2  =  2 .

So " 0  +  2(0)  +  2 " is literally   +   +  . The whole point of choosing x =  -2  and y =  1  is that those values force both squares to  0  - the smallest a square can ever be - leaving only the constant  2  behind. That's why  2  is the minimum.

Answer: C

If x and y are real numbers, then the minimum value of x^2+4xy+6y^2-4y

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If x and y are real numbers, then the minimum value of x^2+4xy+6y^2-4y

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