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. 4
• E. None of the above

Hint:

Always try completing the square when finding minima of quadratic expressions in two variables.

Answer 1

The Correct Option is C

Step 1: Write the expression.
We are given $E = x^2 + 4xy + 6y^2 - 4y + 4$.
Step 2: Try grouping.
$E = (x^2 + 4xy + 4y^2) + (2y^2 - 4y + 4)$.
Step 3: Simplify.
$E = (x + 2y)^2 + (2y^2 - 4y + 4)$.
Now, $2y^2 - 4y + 4 = 2(y^2 - 2y + 2) = 2[(y - 1)^2 + 1]$.
So, $E = (x + 2y)^2 + 2[(y - 1)^2 + 1]$.
Step 4: Identify minimum.
Both terms $(x + 2y)^2$ and $(y - 1)^2$ are always non-negative. The minimum values are achieved when $(x + 2y) = 0$ and $(y - 1) = 0$. At $y = 1$, $x = -2y = -2$. Substituting: $E = 0 + 2(0 + 1) = 2$. Step 5: Conclude.
The minimum possible value of $E$ is 2. \[ \boxed{2} \]