Question

If $x$ and $y$ are real numbers, the least possible value of the expression \[ 4(x-2)^2 + 4(y-3)^2 - 2(x-3)^2 \] is:

• A. -8
• B. -4
• C. -2
• D. 0
• E. 2

Hint:

In optimization problems with quadratic expressions, completing the square is the fastest method to identify the minimum value. Always isolate squared terms and constants carefully.

Answer 1

The Correct Option is B

Step 1: Expand the expression. \[ E = 4(x-2)^2 + 4(y-3)^2 - 2(x-3)^2 \] Expand each term: \[ (x-2)^2 = x^2 - 4x + 4, \quad (y-3)^2 = y^2 - 6y + 9, \quad (x-3)^2 = x^2 - 6x + 9 \]

Step 2: Substitute expansions. \[ E = 4(x^2 - 4x + 4) + 4(y^2 - 6y + 9) - 2(x^2 - 6x + 9) \] \[ E = (4x^2 - 16x + 16) + (4y^2 - 24y + 36) - (2x^2 - 12x + 18) \]

Step 3: Simplify. \[ E = 4x^2 - 16x + 16 + 4y^2 - 24y + 36 - 2x^2 + 12x - 18 \] \[ E = (4x^2 - 2x^2) + (-16x + 12x) + (16 - 18) + 4y^2 - 24y + 36 \] \[ E = 2x^2 - 4x - 2 + 4y^2 - 24y + 36 \] \[ E = 2x^2 - 4x + 4y^2 - 24y + 34 \]

Step 4: Complete the squares. For \(x\): \[ 2x^2 - 4x = 2(x^2 - 2x) = 2[(x-1)^2 - 1] = 2(x-1)^2 - 2 \] For \(y\): \[ 4y^2 - 24y = 4(y^2 - 6y) = 4[(y-3)^2 - 9] = 4(y-3)^2 - 36 \] So, \[ E = [2(x-1)^2 - 2] + [4(y-3)^2 - 36] + 34 \] \[ E = 2(x-1)^2 + 4(y-3)^2 - 4 \]

Step 5: Minimize. Since squares are always non-negative: \[ E_{\min} = -4 \quad \text{(when $x=1$, $y=3$).} \]

Step 6: Conclusion. The least possible value is \(-4\).

Final Answer: \[ \boxed{\text{B. -4}} \]